Does every metrizable space have point-countable base? Just as the title explains: Does every metrizable space have point-countable base? Thanks ahead:)
 A: Yes. This is a standard consequence of A. H. Stone's theorem that every open cover of a metrizable space has a locally finite and $\sigma$-discrete open refinement. This implies that every metrizable space has a $\sigma$-locally finite base, i.e. a base $\mathcal{B} = \bigcup_{n=1}^\infty \mathcal{B}_n$ such that each $\mathcal{B}_n$ is locally finite.
The idea is to extract for each $n$ a locally finite open refinement $\mathcal{B_n}$ of the collection of all balls of radius $1/n$ and to show that $\mathcal{B} = \bigcup_{n=1}^\infty \mathcal{B}_n$ is a base. Indeed: if $B(x,r)$ is a ball and $n$ is so large that $\frac{2}{n} \lt r$ then every element $U \in \mathcal B_n$ containing $x$ must be contained in $B(x,r)$ since $\operatorname{diam}(U) \leq 2/n$. Clearly, every point is contained in at most countably many elements of $\mathcal{B}$.
See e.g. Engelking, Theorem 4.4.1 and Corollary 4.4.4 or Munkres, Step 2 of the proof of the Nagata-Smirnov metrization theorem 40.3 and Lemma 39.2 for more details.
