$A$ closed and $B$ compact $\implies \exists U$ open convex and balanced with $(A + U) \cap (B + U) = \emptyset$

Question

Let $X$ be a topological vector space, a set $U \subset X$ is said to be balanced if $\lambda u \in U$ for any $u \in U$ and $\lambda \in [-1,1]$

I would like to prove the following theorem

Theorem Let $X$ be a topological locally convex vector space, let $A \subset X$ closed, $B \subset X$ compact such that $A \cap B = \emptyset$, then there exists a open convex balanced set $U$ containing $0$ such that $(A + U) \cap (B + U) = \emptyset$

This is the proof of the theorem when $X$ has a norm $||\,||$

Proof

Suppose by contradiction that $(A + B(0,\epsilon) ) \cap (B + B(0,\epsilon) ) \neq \emptyset$ for any $\epsilon > 0$. Let $\epsilon_n \downarrow 0$, let $x_n \in (A + B(0,\epsilon_n) ) \cap (B + B(0,\epsilon) )$ Then there exists $a_n \in A$, $b_n \in B$, $y_n,z_n \in B(0,\epsilon_n)$ such that

$x_n = a_n + y_n = b_n + z_n$

Since $B$ is compact $(b_n)$ as a convergent subsequence, WLOG I can assume that $(b_n)$ itself is convergent.

Since $a_n = b_n + (z_n - y_n)$ and $||z_n - y_n|| \leq ||z_n|| + ||y_n|| \leq 2\epsilon_n$ I have that $(a_n - b_n)$ converges to $0$, therefore $a_n$ is convergent and it's limit is the same of $(b_n)$, but, since the limit of $(a_n)$ is in $A$ ($A$ is closed ) and the limit of $(b_n)$ is in $B$ I have that $A \cap B$ is non-empty, a contradiction.

I tried to generalize this proof to the general case, this is what I came up with :

Is the proof actually correct? If it is, can it be generalized to the case where $X$ is not $T2$?If it's not, why? and is there a way to "fix" the mistakes in the proof?