I saw the following statement in some notes without proof, and I have been trying to verify it (using finite open covers and triangle inequalities) without success, so I would like to ask for some suggestions.
Let $X$ be a compact metric space. For a subset $Y$ of $X$ and $\varepsilon>0$, let $N_\varepsilon(Y)=\bigcup_{y\in Y}B_\varepsilon(y)$, where $B_\varepsilon(y)$ denotes the open ball of radius $\varepsilon$ about $y$. Let $A$ and $B$ be closed subsets of $X$. Then for any $\varepsilon>0$, there exists $\delta>0$ such that $N_\delta(A)\cap N_\delta(B)\subseteq N_\varepsilon(A\cap B)$.
Best Answer
Let's set up some nested family of compact sets: $$ E_k = \overline{N_{1/k}(A)\cap N_{1/k}(B)} \setminus N_\varepsilon(A\cap B) $$ We want to show some $E_k$ is empty. If not, then there exists $x\in \bigcap_k E_k$. By the construction of $E_k$ we have $\operatorname{dist}(x, A)\le 1/k$ for every $k$, hence $\operatorname{dist}(x, A)=0$, hence $x\in A$ (here we use the fact that $A$ is closed). Similarly $x\in B$. But then $x\in A\cap B$ which is definitely disjoint from any $E_k$ by construction. A contradiction.
Just as an aside: the property of being closed is important: the statement would fail for $A=[0, 1]$ and $B=(1, 2]$ on the real line.