Suppose $f:X\to Y$ is continuous at $a$. Can I use $a$ in my $\epsilon-\delta$ proof? For example, is the following proof valid?
Assume $f:\mathbb{R}\to\mathbb{R}$ is continuous everywhere. Let $a\in\mathbb{R}$ be a designated point of continuity. Let $\epsilon>0$ and find $\delta>0$ such that $|x-a|<\delta$ implies $|f(x)-f(a)|<\min(\epsilon,\frac{\epsilon}{\epsilon+2|f(a)|})$, then
\begin{align*}
|f(x)-f(a)|&<\frac{\epsilon}{|f(x)|+|f(a)|}\hspace{15 mm} b/c\hspace{5 mm} |f(x)|<\epsilon+|f(a)|\\
&\leq\frac{\epsilon}{|f(x)+f(a)|}
\end{align*}
Therefore, $|f^2(x)-f^2(a)|<\epsilon$ and so $f^2$ is continuous on $\mathbb{R}$.
I believe this is good because $a$ is chosen arbitrarily from the outset. I just don't like it. If it is indeed valid, maybe you can help convince me it's not so bad?
 A: Yes, your proof is fine, although your approach seems a little bit convoluted. What I mean by that is that one would start estimating the quantity $\vert f^2(x) - f^2(a) \vert$ and only then see what constants you need to get the correct $\delta-\varepsilon$ estimate. For example, I would approach this problem in the following way. We have that $\vert f^2(x) - f^2(a) \vert = \vert f(x) - f(a) \vert \vert f(x) + f(a) \vert \leq \vert f(x) - f(a) \vert (\vert f(x) \vert + \vert f(a) \vert).$ Let $0 < \varepsilon < 1.$ Since $f$ is continuous at $a$, let $\delta_\varepsilon > 0$ such that $\vert f(x) - f(a) \vert \leq \min\{\frac{\varepsilon}{2 \vert f(a) \vert + 1}, 1\}$ whenever $\vert x - a \vert < \delta_\varepsilon$. Then from the above estimate we infer that $\vert f^2(x) - f^2(a) \vert \leq \frac{\varepsilon}{2 \vert f(a) \vert + 1} (2 \vert f(a) \vert + 1) = \varepsilon$ whenever $\vert x - a \vert < \delta_\varepsilon.$ Therefore, $f^2(\cdot)$ is continuous.
Note that as I said in the beginning of the answer, your proof is also correct, so I am not debating it, rather just sharing what is in my opinion a more natural way to get your message across. After all, it's just a matter of taste. Finally, in regards to your original question, when you are interested in proving the continuity of a function $f$ at a certain point $a,$ you are allowed to use the point $a$ when choosing the $\delta_\varepsilon.$ Similarly, you may use it to assert continuity by replacing $\varepsilon$ with a function involving $\varepsilon$ and $a$ as long as you make sure that you can make the desired bound go to $0$ as this is your ultimate goal. That is because $a$ and $\varepsilon$ are both fixed, so you can play around with them to make your proof look neat. However, if you are trying to prove uniform continuity of a function for example, you are not allowed to use any points in your argument since you are interested in a uniform distance between any two points that allow you to make the distance between their images arbitrarily small. I hope this helps. :)
