Vtyjkyuk

I am prooving a theorem where there is a statement saying if X is a separable F-space( I.e. a topological vector space with a complete translation invariant metric) it has no isolated point.
I am unable to prove this statement so that I can proceed further. Kindly help me understanding the statement.
Thank you for your time and effort

If $x$ is an isolated point then there exists $r>0$ such that $x=B(x,r)$. By translation invariance of the metric it follows that $y=y-x+x=y-x+B(x,r)=B(y,r)$ for any $y$. It follows that $y$ is open for every $y$. This contradicts separability. [ Write the whole space as a union of singletons. In a separable space any union of open sets is a countable union. This makes the entire space countable but no vector space other than $0$ is countable].

Well, this is false--the trivial vector space $0$ has an isolated point. But more generally, it is true that no nontrivial topological vector space $X$ has an isolated point (no need to assume it is a separable F-space). This is immediate from the continuity of scalar multiplication: for any $xin X$, $f(t)=tx$ is a continuous map $[0,1]to X$ which gives a path from $0$ to $x$, so $X$ is path-connected.