We know that the function $f(x)=x^{2}$ is continuous at $a$ for all $a$. In other words, if $a$ is any number, then for every $\epsilon > 0$ there is some $\delta>0$ such that, for all $x$, if $|x-a|<\delta$, then $|x^{2}-a^{2}|<\epsilon$.
Of course, $\delta$ depends on $\epsilon$. But $\delta$ also depends on $a$-the $\delta$ that works at $a$ might not work at $b$. Indeed, it's clear that given $\epsilon>0$ there is no one $\delta >0$ that works for all $a$, or even for all positive $a$. In fact, the number $a+\delta/2$ will certainly satisfy $|x-a|<\delta$, but if $a>0$, then
$\biggl|(a+\dfrac{\delta}{2})^{2}-a^{2}\biggr|=\biggl|a\delta + \dfrac{\delta^{2}}{4}\biggr|\ge a\delta$,
and this won't be $< \epsilon$ once $a > \epsilon /\delta$. (This is just an admittedly confusing computational way of saying that $f$ is growing faster and faster!)
On the other hand, for any $\epsilon > 0$ there will be one $\delta >0$ that works for all $a$ in any interval $[-N,N].$ In fact, the $\delta$ which works at $N$ or $-N$ will also work everywhere else in the interval.
So when considering the function $f(x)=x^{2}$, I see how for any $\epsilon >0$, $\delta = \dfrac{\epsilon}{2N}$ works for all $a$ in any interval $[-N,N]$. Now, how would one go about showing that the $\delta$ which works at $N$ or $-N$ also works everywhere else in the interval?
Intuitively, the $\delta$ that works at $a$ is $\frac \epsilon{|f'(a)|}$. As $f'(x)=2x, |f'(x)|$ is increasing away from $0$ so the smallest $\delta$ comes at the furthest point from $0$, which is $\pm N$. The fact that you can find such a $\delta$ says $f(x)=x^2$ is uniformly continuous over the interval. This is important, as it is an assumption in many powerful theorems.
To prove it takes some care. In fact, $\delta=\frac \epsilon{2N}$ would not be quite small enough if $x$ were allowed to go outside $[-N,N]$. Let $N=1, \epsilon=0.01$. Then $f(1)=1$ and you would claim $\delta=\frac \epsilon{f'(1)}=\frac {0.01}2=0.005$ works, but $f(1.005)-f(1)=1.010025-1=0.010025 \gt 0.01$. The second degree term in the Taylor series kills you.
Aside from that detail, you can say let $a \in [-N,N]$. Then $|f(a+\delta)-f(a)|=|(a+\delta)^2-a^2|=|2a\delta+\delta^2|\lt |2a\delta|+\delta^2\le 2N\delta+\delta^2$ so if $\delta$ works at $N$ it works at $a$.