Proof that $f(x) = x^2$ is continuous ($\delta-\epsilon$)? [duplicate]

You have probably seen an $\epsilon$-$\delta$ proof before, so writing a basic proof won't tell you anything new. So I will write how I think when I prove something like this. First off, the definition:

$f(x) = x^2$ is continuous at $x_0$ if, for any $\epsilon>0$, we can find a $\delta>0$ such that for any $c$ with $|x_0-c|<\delta$, we have $|f(x_0)-f(c)|<\epsilon$. $f$ is continuous if it is continuous at every real number.

In terms of how we write the proof, that means we're given arbitrary $x_0, \epsilon$, and we must find a $\delta$ that works. You have already deduced that $|f(x_0)-f(c)| = |x_0^2 - c^2| = |x_0-c|\cdot |x_0+c|$, which is good. We will need that. Why will we need that? Because the first factor of that factorisation is the one thing we have real control over: it will be smaller than $\delta$, and we are free to choose $\delta$. We want the entire expression to be smaller than whatever $\epsilon$ we're given, so the only thing left is to make sure that the second factor, $|x_0+c|$ doesn't ruin things too much.

The standard thing to do to gain control over that term is to declare that whatever happens, I will never choose a $\delta$ which is larger than $1$ (nothing special about $1$ here, I might as well choose $1000$ or $\pi$, but I like $1$). That forces $c$ to be rather close to $x_0$, which means explicitly that $|x_0+c|<2|x_0| + 1$.

Thus, as long as we vow to never pick $\delta$ larger than $1$, that gives us $$ |f(x_0)-f(c)| = |x_0^2 - c^2| = |x_0-c|\cdot |x_0+c|<\delta(2|x_0| + 1)\tag{*} $$ Note that we haven't really picked a $\delta$ yet, we just know that whichever one we pick, as long as it's smaller than or equal to $1$, the inequality $\text{(*)}$ holds, which is good. Why is that good? Because ultimately, we want to be able to choose a $\delta$ so that the left-hand side is smaller than the $\epsilon$ we're given. This is easily achieved by picking a $\delta$ that makes the right-hand side of $\text{(*)}$ no bigger than $\epsilon$.

Thus we have $$ \begin{align} \epsilon &\geq \delta(2|x_0| + 1)\\ \delta&\leq\frac{\epsilon}{2|x_0| + 1}\tag{**}\end{align} $$ So, which $\delta$ do we actually pick? Well, we vowed never to choose a $\delta$ greater than $1$. While under that vow, we found that any $\delta$ which fulfills $\text{(**)}$ works. Thus, if we pick exactly $\frac{\epsilon}{2|x_0| + 1}$, as long as that's smaller than $1$, and we pick $1$ if $\frac{\epsilon}{2|x_0| + 1}>1$, then we really have fulfilled both requirements, and we are done.

Explicitly, this means that we have picked $\delta = \min\left(1, \dfrac{\epsilon}{2|x_0| + 1}\right)$.