Let $V$ be an abstract ($T_0$) topological vector space over topological field $K$ (We may assume that $K = \mathbb{C}$ or $K = \mathbb{R}$ for simplicity). This means that the only thing that we are allowed to use is that scalar multiplication and addition are jointly continuous in both arguments (and also the $T_0$ properties). The problem is to prove:
For each point $x$ and closed set $Y$ such that $x \not \in Y$ there are exist disjoint open neighborhoods $O$ and $O'$ such that $x \in O$ and $Y \subset O'$.
I will write in the sequel for such sets $O \in \mathcal{U}_V(x),O' \in \mathcal{U}_V(Y)$.
If $U \in \mathcal{U}_V(y)$ then we can squeeze it by factor of $r \in K$ $$ r \ast_y U = \big(rU + \{(1 - r)y\}\big) \cap U $$ which is open neighborhood of $y$.
Proof may start with :
for all $y \in Y$ there must exist $U_y \in \mathcal{U}_V(y)$ such that $x \not \in U_y$, otherwise $x \in \lim_{n \to \infty} y $ and so $x \in Y$ which is not true. Then, $U = \bigcup_{y \in Y} U_y \in \mathcal{U}_V(Y)$ do not contain $x$. In case $x \not \in \delta U$ just let $O' = U$. Otherwise, squeeze all open sets by some values $r_y$: $$ O' = \bigcup_{y \in Y} r_y \ast_y U_y $$ Then $O = \Big(\overline{O'}\Big)^\complement$. How to select $r_y$? Will this approach work?
This is a version of the proof redrawn from topological groups:
$f(x,y) = x - y$ is continues function.
Without loss of generality we may assume that $x = 0$.
Then, select $U \in \mathcal{U}_V(Y) $ such that $0 \not \in U$, then where must exist two open sets $O$ and $O'$ such that $ Y \times \{ 0\} \subset O' \times O \subset f^{-1}$(U) as $y - 0 = y$ for all $y \in Y$.
Now we show that $ A =O' \cap (V \setminus U) + O = \emptyset$.
For every $y \in A$, there is a representation $y = v +z$ for some $z \not \in U$ and $v\in O$ . We also know that $y \in O'$. This means that $y - v \in U$ but we also know that $y - v = z \not \in U$, a contradiction. This shows that $A = \emptyset$
If $U$ was constructed as in the question above, then $O \subset (V\setminus U) + O$ as $0 \not \in U$ . Thus, $O \cap O' = \emptyset$
$\square$
This answer was written in order to close the question as it was essentially answered in the comments.