When proving that $x^2$ is continuous we usually show that $\lim_{x \to\ a} x^2 = a^2$ for any x.
So we show that for any $\epsilon > 0$, there is a $\delta > 0$ such that $0< | x-a | <\delta$ implies that $|x^2-a^2|<\epsilon$. We have $|x^2-a^2| = (|x+a|)(|x-a|) < \epsilon $. At this point the book I'm reading puts a bound on the $|x-a|$ term. Why does this work? Is it because $|x-a|$ can be as small as possible since continuity is a local property, so x can be as close a as we like?
Intuitively, you have the right idea. For $f(x) = x^2$, the bound $|x-a|<1$ is common. It makes sense; why bother considering $x$ more than a unit away from $a$? We only care about $x$ in the immediate vicinity of $a$. However, we need a formal justification.
Let us only consider $x$ such that $|x-a|<1$. Then we have $|x| < 1 + |a|$ so that $|x-a||x+a| \leq |x-a|(|x| + |a|) < |x-a|(2|a| + 1) < \epsilon$. Thus, assuming $|x-a|<1$, we choose
$$\delta_0 = \frac{\epsilon}{2|a| + 1}$$
Now, given that $|x-a|<1$, $|x-a|<\delta_0 \Longrightarrow |x^2 - a^2|< \epsilon$.
Now, you ask, what if $|x-a|>1$. Well, we can avoid this scenario entirely by choosing.
$$\delta = \min\left(\delta_0, 1\right)$$
With such $\delta$, $|x-a|$ satisfies both $|x-a|<1$ and $|x-a|<\delta_0$
Since we have already established that $|x-a| < \delta_0$ and $|x-a|<1$ implies $|x^2 - a^2| < \epsilon$, we are done.