Let $X$ be a convex set in the real Hilbert space with countable basis, $0\notin X$. Is it true that there is a hyperplane containing $0$ and disjoint with $X$?
For an open $X$ this would be a direct consequence of the Hahn-Banach separation theorem.
In the plane, let $$ X = \{(x,y): x>0\} \cup \{(0,y):y>0\} . $$ Note $X$ is convex, and $(0,0) \notin X$. Now try to find a line through $(0,0)$ disjoint from $X$.
What you can find is the line $x=0$, such that $X$ is in the half-plane $x \ge 0$ and $(0,0)$ is in the opposite half-plane $x \le 0$.