How to prove continuity of $e^x$.
Fix $a>1$ (the case $a=1$ is trivial, and $0<a<1$ can be handled by switching to $a^{-1}>1$). I'll assume standard facts about $n\mapsto a^n$ for $n\in\Bbb Z$ are known (it is increasing and satisfies the law of exponents). Now to argue the continuity of $\frac bc\mapsto a^{b/c}=(\sqrt[c]a)^b$ on the set of rational numbers $\frac bc$, which is dense in$~\Bbb R$, one can proceed by the following steps.
$a^{b/c}$ is well defined, namely independent of the representation of the rational number $\frac bc$. This means that whenever $\frac bc=\frac pq$, which by definition means $bq=cp\neq0$, one should have $(\sqrt[c]a)^b=(\sqrt[q]a)^p$. Taking positive integer powers is an injective operation ($x^n=y^n$ implies $x=y$), so we raise both sides to the power $cq$ (assumed positive), so that we must prove $(\sqrt[c]a)^{bcq}=(\sqrt[q]a)^{pcq}$; now we can cancel each root against part of the exponent, and there remains to prove $a^{bq}=a^{cp}$, which is clear.
The function $\frac bc\mapsto a^{b/c}$ is strictly increasing: if $\frac bc<\frac pq$ then $a^{b/c}<a^{p/q}$. Given the previous point, one may assume the fractions have a common denominator $c=q$. Raising both sides to this power (which is also a strictly increasing operation) means it suffices to show $a^b<a^p$; again this is clear.
This increasing function has no jump discontinuities*: for any nonempty open interval$~I$ of positive real numbers, there is some $\frac bc$ with $a^{b/c}\in I$ (a jump discontinuity would give such interval without images). If $I=(x,y)$ put $\gamma=y/x\in\Bbb R_{>1}$, and let $n\in\Bbb N$ be sufficiently large that $\gamma^n>a$. Then $y^n>ax^n$ so there is some integer power $a^m$ of $a$ with $x^n<a^m<y^n$. Taking $n$-th roots (again a strictly increasing operation) it follows that one has $x<a^{m/n}<y$, in other words $a^{m/n}\in I$.
Any increasing function$~f$ defined on a dense set and without jump discontinuities is continuous. Given $x$ in the domain of $f$, and $\epsilon>0$, find values $x_<,x_>$ with $f(x_<)\in(f(x)-\epsilon,f(x))$ and $f(x_>)\in(f(x),f(x)+\epsilon)$ by point$~$3, and take $\delta=\min(x-x_<,x_>-x)$. Then for all $x'$ with $|x-x'|<\delta$ one has $x'\in(x_<,x_>)$ so $f(x')\in(f(x_<),f(x_>))$ by the increasing property of$~f$, and $(f(x_<),f(x_>))\subset(f(x)-\epsilon,f(x)+\epsilon)$, so that $|f(x)-f(x')|<\epsilon$.
This does not yet entirely answer your question, for which it is necessary to show that any function like in 4. can be uniquely interpolated to a continuous increasing function defined on $\Bbb R$. For $x_0$ not in the domain $D$ of $f$ (here an irrational number) one wants to have $f(x_0)=\lim_{x\to x_0, x\in D}f(x)$. To show that the limit exists, one uses that $\sup\{\,f(x)\mid x\in D_{<x_0}\,\}=\inf\{\,f(x)\mid x\in D_{>x_0}\,\}$ (by the absence of jump discontinuities) which is going to be the value of the limit; now the proof that the limit converges can be done by finding values in $D$ sufficiently close to the limit value and some easy but boring estimations, which I will not do here (as per the headache-avoiding request). Also the proof that the extended function is still is continuous and increasing is straightforward.
*This terminology is is not quite proper, as the function on $\Bbb Q \to\Bbb R$ sending $x\mapsto x+[x^3>2]$ (the brackets meaning $1$ if the condition holds, $0$ otherwise) is continuous on its domain, so it is somewhat unfair to accuse it of having a jump discontinuity, as the above implies. The proper term for what I call "function without jump discontinuities" would be something like "function the closure of whose image is connected" (assuming an increasing function defined on a dense set in both cases), not very attractive.
$\mathbf{Result \; 1}$: If $a > 0 $, then $\lim_{n \to \infty } a^{\frac{1}{n}} = 1 $.
If you want a solution to this problem, let me know.
$\mathbf{Result \; 2} $: If $a> 0$, then $f(x) = a^x$ is continuous in $\mathbb{R}$
$\mathbf{Solution}:$ Let $a>0$, and let $\epsilon >0$ .We will employ the sequential characterization of continuity. This means that for any given sequence $(x_n) \subseteq \mathbb{R}$ with $x_n \to x$, then $f(x_n) \to f(x)$. To show this, let $(x_n) \subseteq \mathbb{R}$ be an arbitrary sequence such that $x_n \to x$. Notice by arithmetic,
$$ a^{x_n} - a^x = a^x(a^{x_n-x} - 1) $$
By previous result, we can obtain an $N \in \mathbb{N}$ such that for all $n > N$,
$$ |a^{\frac{1}{n}} - 1 | < \frac{\epsilon}{a^x} $$
Similarly, since $x_n \to x$, we can select $K \in \mathbb{N}$ such that for all $n > K$,
$$ |x_n - x| < \frac{1}{N} $$
Therefore, if we put $M = \max\{N,K\} $, we have
$$ |f(x_n) - f(x) | = |a^{x_n} - a^x| = a^x|a^{x_n-x} - 1| < a^x \frac{ \epsilon}{a^x} = \epsilon$$
Hence, by definition, $f(x_n) \to f(x) $. In particular, $f(x) = a^x$ is continuous as desired.
$\mathbf{Proposition}:$ $\exp(x) $ is continuous on $\mathbb{R}$.
$\mathbf{Solution}:$ Apply $\mathbf{Result \; 2}$ to $a = e$, and we are done.
Consider the function $f:(0,\infty)\to\mathbb R$ such that $f(x)=\int_1^x\frac{\mathrm dt}{t}$. This is continuous, differentiable and, in fact $f'(t)=1/t$ for all $t\in(0,\infty)$. In particular, it is clear that $f'(t)>0$ for all such $t$ so that $f$ is strictly increasing in its domain. Some work will show that $f$ is not bounded neither above not below —indeed, and for example, the first is equivalent to the statement that the integral $\int_1^\infty\frac{\mathrm dt}{t}$ diverges.
Now we have
If $h:(a,b)\to(c,d)$ is a continuous, strictly increasing function which is surjective, then there is an inverse function $h^{-1}:(c,d)\to(a,b)$ and it is itself also continuous and strictly increasing.
This applies to our $f$, and shows that it has an inverse function $f^{-1}:\mathbb R\to(0,\infty)$ which is continuous. Since this is of course the exponential function, we have shown what we wanted.
Later. Notice that one can define the exponential functio in this way —and it is done so classically, too. In any case, It is very easy to check that it is the exponential. Let $h=f^{-1}$. A change of variables shows that $f(xy)=f(x)+f(y)$, so applying inverse functions we see that $h(x+y)=h(x)h(y)$. It follows that $(h(x+y)-h(x))/y=h(x)(h(y)-1)/y$ and, since we know that $h$ is differentiable, it follows that $h'(x)=h'(0)h(x)$ for all $x$. The inverse function theorem allows us to compute $h'(0)=1$, so $h'(x)=h(x)$ and $h(0)=1$. One can now compute the derivative of $h(x)e^{-x}$, see that it is identically zero, adn evaluate his function at $0$ to conclude that $h(x)=e^{-x}$.