Show that $f: \mathbb{R} \to \mathbb{R}: x \mapsto x^2 $ is continuous
My attempt. Let $a \in \mathbb{R}$ be fixed but arbitrary.
Let $\epsilon > 0$. Choose $\delta := \min\{\sqrt{\frac{\epsilon}{2}}, \frac{\epsilon}{4|a|}\}$
Then, for all $x \in \mathbb{R}$ such that $|x-a| <\delta$, we have:
$$|x^2 - a^2| = |x-a||x+a| = |x-a||x-a + 2a| \leq|x-a|(|x-a| + 2|a|) \leq \delta^2 + 2\delta|a|\leq \epsilon$$
Hence, $f$ is continuous at $a$ and because $a$ is arbitrary it follows that $f$ is continuous on its domain, as desired.
Is this correct? (not a duplicate btw, I have my own proof with another choice of delta than other proofs)
This proof looks good. The only issue I can see is that your definition of $\delta$ has an undefined expression in it if $a=0$. This should be pretty easy to fix, though.
Depending on what your instructor is looking for, you might want to make explicit why the two terms in your penultimate expression are each less than $\frac{\epsilon}{2}$