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 $x, \epsilon, a$.

For any $\epsilon\gt0$, and $x\in A$, let $\delta=\min\left(\frac{|x-a|}2,\frac{|x-a|^2}2\epsilon\right)$. If $|y-x|\le\delta$, then
$$
\begin{align}
|y-a|
&\ge|x-a|-|y-x|\tag1\\[3pt]
&\ge|x-a|-\frac{|x-a|}2\tag2\\
&=\frac{|x-a|}2\tag3
\end{align}
$$
Explanation:
$(1)$: triangle inequality
$(2)$: $|y-x|\le\delta\le\frac{|x-a|}2$
$(3)$: arithmetic
Therefore,
$$
\begin{align}
\left|\frac1{|x-a|}-\frac1{|y-a|}\right|
&=\frac{|\,|y-a|-|x-a|\,|}{|x-a||y-a|}\tag4\\
&\le\frac{|y-x|}{|x-a|\frac{|x-a|}2}\tag5\\
&\le\epsilon\tag6
\end{align}
$$
Explanation:
$(4)$: arithmetic
$(5)$: triangle inequality and $(3)$
$(6)$: $|y-x|\le\delta\le\frac{|x-a|^2}2\epsilon$
This says that $\frac1{|x-a|}$ is continuous at any $x\in A$.
