# Arbitrary basis of a separable metric space

Every separable metric space has a countable basis. However, is it true that given ANY basis for the metric space, we can produce an arbitrary open set using only countably many of the basis elements? A proof for this would maybe look something like this: Take an arbitrary open set $X$, then consider the intersection of $X$ with our dense countable set of points $Y$. Let this (countable) intersection $X \cap Y = S = S_1,S_2, \dots$. Supposing we already have generated $X$ using basis elements $(B_i)_{i\in I}$, we can then choose $B_1,B_2,\dots$ that contain the (countable) set of points $S_1,S_2,\dots$ respectively, and we can use the open-set property of the $B_i$ to say that a neighborhood around each $S_i$ must be contained in the $B_i$ also. But now I get stuck, because I can't control the sizes of these neighborhoods, and it could be that the subsequence of the $S_i$ that converges to a given point in $X\backslash B$ could have decreasing neighborhood sizes and never actually contain the limit point. How do I fix this proof, or is the claim just false?

The weight $w(X)$ of a topological space $X$ is the smallest cardinality (infinite) of a base for the space $X$. It is proved here Smallest/Minimal bases of a topological space that every base of $X$ contains a subset with cardinality $w(X)$ that is also a base for $X$.
If $X$ is second countable, then $w(X)=\aleph_0$. So, every base $\mathcal{B}$ for $X$ contains a countable subset that is also a base. Hence every open set in $X$ is union of countable many sets of $\mathcal{B}$.