First year calculus student: why isn't the derivative the slope of a secant line with an infinitesimally small distance separating the points?

Because from my understanding, in order for it to be a tangent line, it intersects the curve at one point only, however Δx approaches zero, it never reaches it, so Δx must be greater than zero, however infinitesimally small, correct?

You're right. We don't ever reach that point. We take a limit.

The colloquialism, "reaching the point" is a good anthropomorphic description. Limits allow us to stretch the constraints of the real numbers by pushing towards the infinite and infinitesimal. Technically, though, to venture into such territory, we need to properly define limits. This is often introduced with the epsilon-delta formalization.

Say there exists a limit $f'(x)=\lim_{\Delta x\rightarrow0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$. Then for every $\epsilon>0$, there exists some $\delta>0$ such that whenever $0<\Delta x<\delta$, we find $|f'(x) - \frac{f(x+\Delta x)-f(x)}{\Delta x}|<\epsilon$.

We can heuristically think of the last paragraph as the following: our derivative exists if for every positive number $\epsilon$ and $\delta$, including the most ridiculously small numbers you can ever imagine, whenever $\Delta x$ is trapped between zero and any of these ridiculously small numbers, the difference between our derivative and the original expression is imperceptible.

But wait a minute, you say

...Δx must be greater than zero, however infinitesimally small, correct?

The epsilon-delta definition seems to hint that as well, but there's a catch: $$|f'(x) - \frac{f(x+\Delta x)-f(x)}{\Delta x}|<\epsilon$$

This is not less than some real positive number $\epsilon$. This is less than ANY POSSIBLE real positive number $\epsilon$. Such a concept only exists within the formalism of a limit, and is by no means a measurable quantity. That's what is meant by infinitesimal.

Due to the limit, then, the derivative cannot represent any possible secant line. There are no two points corresponding to $x+\Delta x$ and $x$ that are indistinguishable! The value we reach has converged to that which represents the slope of the tangent.

Added note: $\Delta x\rightarrow 0$ doesn't just imply that $\Delta x$ is running through the positive numbers towards zero. For the limit to exist, we typically require it to be two-sided, meaning that $\Delta x\rightarrow0^+$ and $\Delta x\rightarrow0^-$ must produce the same result. In either case, the difference between $\Delta x$ and zero becomes vanishingly small.

You're treating the notion of "tangent" as pre-existing and asking how we can say that the derivative is (the slope of) the tangent. But the situation is more complex than that – by defining the derivative, we're defining what we mean by tangent in many cases in which there was no such pre-existing concept. So the task is to clarify the pre-existing, more restricted concept of "tangent" and check whether it coincides with the one induced by the derivative in the cases where it applies.

Your description of the tangent as intersecting the curve at one point doesn't capture the pre-existing concept of tangent. Let's consider four examples:

a) $f(x)=x^3$ at $x=1$. The tangent coincides with the curve at two points (at $x=1$ and at some negative $x$).

b) $f(x)=x$ at $x=0$. The tangent coincides with the curve everywhere. And what's worse, any other line through the origin intersects the curve only once – so by your definition the tangent is the only non-tangent.

c) $f(x)=x^3$ at $x=0$. Here the tangent indeed coincides with the curve in one point only, but the curve lies on different sides of it, and it's at least not obvious how to distinguish this from an ordinary line intersecting a curve without using calculus.

d) $f(x)=x^2\sin\frac1x$ at $x=0$. Here the tangent as defined by calculus (the $x$-axis) intersects the curve infinitely often in any neighbourhood of the origin. I don't know of any way of defining this as the tangent without calculus.

So in order to make your question more precise, you'd need to provide a clear definition of how "tangent" was used before the advent of calculus and what cases it applied to, and then we could check whether the calculus definition coincides with this (and expands it).

I like your starting line "I'm having trouble with the limit approach to calculus ever since I heard about the infinitesimal definition." +1 for that. The short answer to your question lies in that line because of the following non-mathematical theorem:

A non-mathematical theorem: There is no sound theory of infinitesimals available for a beginner in calculus and hence any approach based on infinitesimals is bound to cause confusion.

The fundamental problem with an infinitesimal is that it tries to describe something which is smaller than any positive number and not yet zero. No such thing exists in the real number system. There is a way to define "infinitesimals in sound manner" which is called "Non-standard Analysis" but it is not suitable for a beginner learning calculus.

It is somewhat ironical but also surprising that the rebirth of calculus happened with the idea of infinitesimals and many mathematicians faced similar confusion as OP and therefore there was a concerted effort by the mathematical community to discard the infinitesimals totally and present calculus in a manner which is sound and therefore far less confusing. There was another problem with calculus apart from the infinitesimals and it was that many important and deep theorems had no proofs and this last problem was fixed by a proper theory of real numbers.

In the previous paragraph I mentioned the "rebirth of calculus" because the birth of calculus happened long back in time of Archimedes and the Greeks were the ones who had developed calculus (mainly integral calculus in its basics) without infinitesimals and they also had a proper theory of real numbers. Unfortunately they did not have the differential calculus stuff which was championed so much by the likes of Newton and Leibniz that the very pure theory of calculus was forgotten for a long time.

It is our great luck that the infinitesimal approach has been discarded in our time. The definition of limit which you should focus on is the one involving $\epsilon, \delta$ because they are far less confusing if presented in proper manner. Also some calculus textbooks try to teach limits via jumping on to the notion of derivatives. A limit involved in the concept of a derivative is a "difficult type of limit" (namely the "indeterminate form" $0/0$) and it is better to avoid such an approach. A beginner is well off if he/she studies limit without knowing anything about derivatives and then later learn special kinds of limits which are called derivatives.

Anyway lets come back to tangents, secants, and derivatives. The geometrical concept of a secant is defined for any kind of curve without the need for calculus, but the notion of tangent to a point on a curve can not be defined in purely geometric manner without using calculus concepts. The concept of tangent does not exists apriori rather it is defined in terms of derivative. One exception is that a tangent to a circle can be defined without any appeal to calculus by simply defining it to be a line perpendicular to the radius at the point under consideration.

The idea of a secant line becoming a tangent as the points of secant come infinitely close to each other is a totally non-rigorous idea and it is used by instructors to somehow provide a geometric interpretation for the concept of derivatives. There is nothing like "infinitely" close. Two points are either same or at some specific distance to each other. A secant is line which essentially requires two points on curve. A tangent normally deals with one point on the curve. There is thus no way a secant can become a tangent.

What calculus has to offer here is the following. Let $f$ be a function defined on a certain interval $I$ and let $C$ be the curve which is the graph of the function and let $c$ be any specific point in interval $I$. Let the point $P$ on the curve $C$ be $(c, f(c))$. Consider another point $x \in I$ such that $x \neq c$ and let $Q = (x, f(x))$ be the corresponding point on the curve. The equation of the secant $PQ$ is given by $$Y - f(c) = \frac{f(x) - f(x)}{x - c}(X - c)$$ The slope of this secant $PQ$ is $$\frac{f(x) - f(x)}{x - c}$$ and sometimes for some functions $f$ it is possible that the limit $$\lim_{x \to c}\frac{f(x) - f(x)}{x - c} = f'(c)$$ exists. When this happens we say that the curve $C$ possesses a tangent at point $P$ and its equation is $$Y - f(c) = f'(c)(X - c)$$ There are some corner cases to consider when the above limit is $\pm\infty$ but apart from that the above constitutes the definition of a tangent to a curve at a point $P$.