Let $ B \subseteq E \subseteq\mathbb{R^n}$, where $B$ is compact relative to $E$, $E$ open relative to $\mathbb{R}^n$. Let $A := \{v \in \mathbb{R}^n \mid d(v,B) \leq \epsilon\} \subseteq E$, where $d = \Vert \cdot \Vert_E$. Prove that $A$ is compact, relative to $E$.
My attempt:
Define $g: \mathbb{R}^n \to \mathbb{R}: v \mapsto d(v,B)$. Then $g$ is continuous, as it is Lipschitz, and $A = g^{-1}((- \infty,\epsilon])$, and it follows that $A$ is closed in $\mathbb{R}^n$, as preimage of a closed set.
Now, since $B$ is compact, it is bounded, and there exists $x \in E, r > 0$ such that $B \subseteq B_E(x,r)$.
Then, let $v \in A$, and let $k \in B$. It follows that:
$$d(v,x) = \inf_{k \in B}d(v,x) \leq \inf_{k \in B} (d(v,k) + d(k,x)) \leq d(v,B) + \inf_{k \in B} d(k,x) \leq d(v,B) + \inf_{k \in B} r \leq \epsilon + r$$
Hence, $A \subseteq B(x, \epsilon + r +1)$ and $A$ is closed and bounded in $\mathbb{R}^n$, so it is compact in $\mathbb{R}^n$ and therefore it is compact in $E$ as well.
Is this correct?
Your proof is correct.
The part to prove that $A$ is closed is great, but I would simplify a little the part to prove that it is bounded.
Indeed since $B$ is bounded you have $B\subset B_E(0,R)$ for a certain $R>0$.
Now if $v\in A$, you have $v\in B_E(0,R+\epsilon)$ by triangular inequality, since
$$d(v,0)\leqslant d(v,B)+d(B,0)\leqslant \epsilon +R.$$
It is quite the same as what you did, but I find it clearer not to use so many "inf".