I'm reading this theorem and its proof.
Theorem: Let $A$ and $B$ be two nonempty disjoint closed subsets of a real (not necessarily Hausdorff) t.v.s. $X$ such that $A$ is compact. Then there exists a neighborhood $V$ of $0$ such that $(A+V) \cap B=\emptyset$.
Proof: For each $a \in A$, fix $U_{a} \in \mathcal{U}(0)$ such that $\left(a+U_{a}\right) \cap B=\emptyset$, and an open $V_{a} \in \mathcal{U}(0)$ with $V_{a}+V_{a} \subset U_{a}$. Since the sets $a+V_{a}(a \in A)$ form an open covering of the compact set $A$, there exists $a_{1}, \ldots, a_{n} \in A$ such that $A \subset \bigcup_{i=1}^{n}\left(a_{n}+V_{a_{i}}\right)$. The set $$ V=\bigcap_{i=1}^{n} V_{a_{i}} $$ is a neighborhood of 0 , and $A+V \subset \bigcup_{i=1}^{n}\left(a_{i}+V_{a_{i}}+V\right) \subset \bigcup_{i=1}^{n}\left(a_{i}+V_{a_{i}}+V_{a_{i}}\right) \subset$ $\bigcup_{i=1}^{n}\left(a_{i}+U_{a_{i}}\right)$. Then $(A+V) \cap B \subset \bigcup_{i=1}^{n}\left(a_{i}+U_{a_{i}}\right) \cap B=\emptyset$.
Below my shorter proof which I found more intuitive. Could you have a check on it?
My attempt: Assume the contrary that for each neighborhood $V$ of $0$, there is $x_V \in (A+V) \cap B$. Let $a_V \in A$ and $y_V \in V$ such that $x_V = a_V + y_V$.
Let $\mathcal V$ be the collection of all neighborhood of $0$. We endow $\mathcal V$ with a partial order $\le$ by $V_1 \le V_2 \iff V_2 \subset V_1$. Clearly, $y_V \to 0$.
Because $A$ is compact, there is a subnet $(a_{\psi(d)})_{d\in D}$ of $(a_V)_{V \in \mathcal V}$ and $\overline a\in A$ such that $a_{\psi(d)} \to \overline a$. Here $\psi:D \to \mathcal V$ is monotone cofinal. It follows from $y_V \to 0$ that$y_{\psi(d)} \to 0$, so $x_{\psi(d)} \to \overline a$. Because $A, B$ are closed, we get $\overline a \in A \cap B$, which is a contradiction.