Struggling with basic limit problem As part of some other proof, I'm trying to show the following:
Let $A \subset \mathbb{R}^n$ and $a \in \overline{A}-A$ (ie. in the closure of A but not in A) so that the function $f : A \to \mathbb{R}$ defined by $f(x) = \frac{1}{|x-a|}$ is well-defined on $A$. I want to show that the function is continuous everywhere on $A$. Now I went through the usual rigmarole which I'll outline below, but essentially I
am struggling to find a way to bound the $\frac{1}{|y-a|}$ that you'll see in the attempted proof below. I seem to have forgotten how to deal with cases like this from single-variable calculus.
Attempt:
Fixing any $x \in A$ and given any $\epsilon$, we want to be able to bound, for any other $y \in A$ (subject to that $y$ being in the ball defined by the $\delta$ we're trying to determine),
$$|f(y)-f(x)|=\left|\frac{1}{|y-a|}-\frac{1}{|x-a|}\right|=$$
$$\frac{1}{|x-a||y-a|}\left||y-a|-|x-a|\right|\leq\frac{1}{|x-a||y-a|}|y-x|$$
so that we have it in the nice form where if we require $\delta$ to be the $min(\epsilon|x-a|, B)$, where $B$ is that something I'm missing to put a bound on the $\frac{1}{|y-a|}$, then we are done. Can someone help point out what I need?
Roughly, it seems clear to me that I need to show $|y-a|>C$ for some $C$ that depends only on $x, \epsilon, a$.

The intuition is that if $y$ is close to $x$, then $|y-a|$ is close to $|x-a|$.
Formally, to obtain $|y-a| > C$ as you mentioned in your last sentence,
$$|y-a| \ge |x-a| - |x-y| \ge |x-a| - \delta.$$
