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From General Topology by Kelley,
Chapter 1, Prob. $\textbf F$, page 57.

Let the topology $\tau$ of space $X$ have a countable base.
We need to show that each base contains a countable subfamily which is also a base.

From definition - A base $\beta$ for topology $\tau$ in space $X$ is a family of sets s.t,

$$\forall \, a\in\tau \quad a=\cup_i\, b_i \qquad b_i\in \beta.$$

Proof
If $\beta$ is a base for $\tau$, then $\tau\ni a=\cup_i\, b_i$ for $b_i \in \beta$.
So for each $a\in\tau$ the union of $b_i$ is in a subfamily of $\beta$.

Edited Proof:
Let $\beta$ be a countable base for topology $\tau$ in $X$.
Then for each $a\in \tau$ $\quad a=\cup_i \{b_i\}$ for some $b_i\in \beta,\; i\in \Bbb N$.
So by creating a family of sets $\cup_i\{b_i\}$ for each $a$, we have subset $\gamma$ which is a base and is made out of elements of $\beta$, hence countable.

flowian
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  • Simple check: You haven't used the countable base in your proof, so how can it be a proof of a consequence of having a countable base? (Also, you mean the family ${b_i}$, not the union of $b_i$, and you should say what $i$ is.) – David Hartley Jun 12 '20 at 10:22
  • @DavidHartley I guess it is not so obvious when reading. I edited the OP. – flowian Jun 12 '20 at 14:08
  • You're still missing the point. $X$ has a countable base, say $\beta$, and another base, $\alpha$, which may not be countable. You have to find a subset $\gamma \subset \alpha$ which is both countable and a base. – David Hartley Jun 12 '20 at 19:44

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Hint : it suffices to show that said $\beta$ your countable basis and given $\alpha$ another basis, you have that every $b \in \beta$ can be written as countable union of elements of $\alpha$.

In general you shall prove that every open set $s$ can be written as countable union of elements of $\alpha$: write $s$ as $s=\cup \alpha'$ with $\alpha' \subseteq \alpha$. Then consider $\beta' = \{b \in \beta : \exists a \in \alpha', b \subseteq a\}$: clearly $\beta' \subseteq \beta$ and $\cup \beta' = s$. For every $b \in \beta'$ choose one $a_b \in \alpha'$ so that $b \subseteq a_b$, then you have $\alpha'':=\{a_b : b \in \beta'\}$ is a countable subset of $\alpha$ such that $\cup \alpha''=s$

Dunnò000
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