Struggling with basic limit problem

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 $x\in A$.

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,

$$|y-a| \ge |x-a| - |x-y| \ge |x-a| - \delta.$$