Let $E$ be a metrizable topological vector space such that is separable, that is, there is a dense countable set in $E$. I want to prove that: if $A \subset E$ is noncountable, then exists a sequence $(x_n)_{n \in \mathbb{N}} \subset A$ such that $x_n \longrightarrow x$, for some $x \in E$.
One idea was to assume that for some countable $ X \subset E $, then every sequence in $ X $ not converges in $E$ and use the result of that question question and arrive at some contradiction. But I was not successful with that idea.
If $E$ is metrizable and separable, then we don't even need to assume that $E$ is a TVS. Since $E$ is metrizable and separable, it is second-countable, which implies that no uncountable subspace of $E$ is discrete (since a subspace of a second-countable space is also second-countable). In particular, if $A\subseteq E$ is uncountable, then $A$ is not discrete and therefore has a limit point $x\in A$. Since $E$ is second-countable, it is also first-countable, so we can take a countable neighborhood basis $\{B_n\}_{n\in\mathbb{N}}$ for $x$ and choose some $x_n\in B_n\cap A\neq\emptyset$ for each $n\in\mathbb{N}$. Then it follows that $x_n\to x$.