In a second countable space any base has a countable subset that is still a base [closed]

There is nothing special about the countable case, it works for all infinite cardinalities. Proof follows Engelking, General Topology.

Theorem Let $X$ be a space and suppose $X$ has a base $\mathcal{M}$ of size $\kappa$, some infinite cardinal. If $\mathscr{B}$ is any base for the topology of $X$, then there exists a subfamily $\mathscr{B}' \subseteq \mathscr{B}$ such that $\left|\mathscr{B}'\right| \le \kappa$ and $\mathscr{B}'$ is still a base for the space $X$.

So for example, if $X$ has a countable base, all other bases of $X$ can be "thinned out" to a countable base, by possibly throwing some sets away.

Proof:

Let $\mathscr{B}$ be any base for $X$.

Then define $$I = \{(M_1, M_2) \in \mathscr{M} \times \mathscr{M}: \exists B \in \mathscr{B}: M_1 \subseteq B \subseteq M_2 \}$$

Note that $\left|I\right| \le \kappa^2 = \kappa$, and we apply the Axiom of Choice to pick for each $i \in I$ (where $i = (M^i_1, M^i_2)$), some $B_i \in \mathscr{B}$ such that $M^i_1 \subseteq B_i \subseteq M^i_2$.

We claim that $\mathscr{B}' := \{B_i: i \in I\} \subseteq \mathscr{B}$ is also a base for $X$, and clearly $\left| \mathscr{B'}\right| \le |I| \le \kappa$ and so we would be finished.

To see it is a base: let $O$ be open in $X$ and $x \in O$. We have to find some $B_i$ that sits between them.

First use that $\mathscr{M}$ is a base and find $M_2 \in \mathscr{M}$ such that

$$x \in M_2 \subseteq O$$

Then use that $\mathscr{B}$ is a base (applied to $x$ and $M_2$) and find $B \in \mathscr{B}$ such that

$$ x \in B \subseteq M_2 \subseteq O$$

Again apply that $\mathscr{M}$ is a base (to $x$ and $B$) and find $M_1 \in \mathscr{M}$ such that

$$ x \in M_1 \subseteq B \subseteq M_2 \subseteq O$$

Aha! We have that $i:= (M_1, M_2) \in I$ (we've forced it that way using the base property ) So we have already picked some $B_i = B_{(M_1, M_2)}\in \mathscr{B}'$ (it's probably some other member, not necessarily our $B$ from above), such that

$$ x \in M_1 =M^i_1 \subseteq B_i \subseteq M^i_2 = M_2 \subseteq O$$

And we have found the required member of $\mathscr{B}'$ between $x$ and $O$. This finishes the proof.

(footnote: a "modern proof" would pick an elementary submodel of a sufficiently large fragment of ZFC of size $\kappa$, a bit of overkill for this case.)