Open set around intersection of two convex sets Consider two closed (compact if needed) convex sets $E$ and $F$. Define the $\epsilon$ neighborhood of set $E$ as
$$E_\epsilon = \cup_{x \in E} B_\epsilon (x),$$ where $B_\epsilon(x)$ is the open $\epsilon$-ball around $x$.
Now, given a $\delta>0$, is there always an $\epsilon>0$, such that
$$E_\epsilon \cap F \subseteq (E\cap F)_\delta.$$

One counter example for $E$ not closed:

The intersection $E\cap F$ is the top right corner, and its $\delta$-neighborhood is just the $\delta$-ball.
However, for any $\epsilon>0$, $E_\epsilon \cap F = F$, which would easily be out of $(E\cap F)_\delta$.

Some backgrounds:
I started with the problem without the closedness and convexity assumption, and intuitively it makes sense. Then I started to see counter examples. With (almost) the strongest assumption, is the above statement correct?
One more assumption I can think of is to assume that $\mathrm{int}(E) \cap \mathrm{int}(F) \ne \emptyset$.
Thanks for any help / suggestions / hints!
 A: Assuming $X=\mathbb{R}^d$, I think the answer is true. Assume towards contradiction that the statement is not true. Then for every $n$ there exists an $x_n\in F$ such that
$$d(x_n,E)<\frac{1}{n} \quad  \text{and} \quad d(x_n,E\cap F)\geq \delta.$$
If $F$ is compact there exists an $x_\infty \in F$ such that $x_n\to x_\infty$. Since $E$ is closed and $d(x_n,E)\to 0$, we have $x_\infty \in E$. Hence, $x_\infty \in E\cap F$ while $d(x_\infty,E\cap F)\geq \delta$. This is clearly a contradiction.
My proof is weird because it just needs for the intersection not to be empty and the sets to be closed. Nowhere do I rely on convexity.
A: If we allow $E$ and $F$ to be compact, and $E \cap F \neq \emptyset$ then the result is quite trivial in any metric space. In fact we don't need compactness - if one of them is bounded statement holds.
If $E$ was bounded, so would be $E_\epsilon$, so $E_\epsilon \cap F$ is bounded either way (as a subset of bounded set). Choose ball $B(x, r) \supseteq E_\epsilon \cap F$ and any point $p \in E\cap F$. For $\delta = r + d(x, p)$ we have $B(x, r) \subseteq B(p, \delta) \subseteq (F\cap E)_\delta$.
So sets don't even need to be closed.
