$\forall\epsilon>0\exists\delta_\epsilon>0: U_{\delta_\epsilon} (A)\cap U_{\delta_\epsilon} (B)\subseteq U_\epsilon(A\cap B) $ Let $(M,d)$ be a metric space, $A$ a closed and $B$ a compact subset with $A\cap B\neq\emptyset$. Define for a set $X\subseteq M$ and $\epsilon>0$ $$U_\epsilon(X):=\{m\in M: d(m,X)<\epsilon\}$$ with $d(m,X):=\inf\{d(m,x): x\in X\}$. It is obvious that for all $\epsilon>0$ $$U_\epsilon(A\cap B)\subseteq U_\epsilon(A)\cap U_\epsilon(B)$$ holds. But holds also a kind of converse statement, that for every $\epsilon>0$ there exists $\delta_\epsilon >0$ so that $$U_{\delta_\epsilon} (A)\cap U_{\delta_\epsilon} (B)\subseteq U_\epsilon(A\cap B) $$ holds true?
If this is not true for every metric space, is it maybe true for $M=\mathbb{R}^n$?
 A: I think this is true.
Let $\varepsilon > 0$. Lets assume that there isn't such a $\delta=\delta(\varepsilon)$. Then for avery $n\in \mathbb{N}$ (and $\delta_n=\frac{1}{n}$), there exists a $x_n\in U_{\frac{1}{n}}(A)\cap U_{\frac{1}{n}}(B)$ but $x_n$ is not an element of $U_{\varepsilon}(A\cap B)$. Now, by definition, $\forall n\in \mathbb{N}$, $\exists a_n\in A$ and $b_n \in B$ such that $d(x_n,a_n), d(x_n,b_n)< \frac{1}{n}$.
Since $B$ is compact there is a subsequence $(b_{k_n})$ of the sequence $(b_n)$ and a $b\in B$ s.t. $d(b_{k_n}, b) \to 0$. Follows that $$d(a_{k_n}, b)<d(a_{k_n}, x_{k_n})+d(x_{k_n}, b_{k_n})+d(b_{k_n}, b) \to 0 \ , \quad as \ n\to\infty$$
and since $A$ is closed, $b\in A$. In other words $b\in A\cap B$.
In the a similar manner we show that $d(x_{k_n}, b) \to 0$. Hence, eventually $x_{k_n}$ lies into the open ball $B(b, \varepsilon)$ and therefore into $U_{\varepsilon}(A\cap B)$. But this is a contradiction of how the sequence $x_n$ was constructed.
P.S. 1
This is not a very intuitive/visual/geometric proof. I would like to see if someone has such a proof. Maybe this way can give more information of how $\delta$ compares to $\varepsilon$, if that's possible.
P.S. 2
I am pretty sure that this does not work without the compactness.
