In a metric space with a countable base, how does every open cover have a countable subcover? Let $X$ be a mertic space, and let $\left\{ V_{\alpha} \right\}$ be a collection of open subsets of $X$ such that, for every $x \in X$ and for every open set $G \subset X$ with $x\in G$, there is some $V_\alpha$ such that
$$x \in V_\alpha \subset G.$$
Then the collection $\left\{ V_\alpha \right\}$ is said to be a base for $X$.
Now suppose $X$ has a countable base $\left\{ V_1, V_2, V_3, \ldots \right\}$, and let $\left\{O_\beta \right\}$ be an open cover of $X$; that is, let $\left\{O_\beta \right\}$ be some collection of open sets such that
$$ X \subset \bigcup_{\beta} O_\beta. $$
Then how to show that some countable subcollection of $\left\{ O_\beta \right\}$ also covers $X$?
Of course, every element $x \in X$ is in some $O_\beta$, which in turn is a union of some subcollection of the countable collection $\left\{V_1, V_2, V_3, \ldots \right\}$.
What next? How to obtain a countable subcollection of $\{O_\beta\}$?
 A: well you need the axiom of choice. since $\{V_\alpha\}$ is countable, and is a basis, for every $V_i$ choose $O_i \in  \{O_{\beta}\}$ satisfying $V_i \subset O_i$. we may write $O_i = \psi(V_i)$
every $x \in X$ is covered by at least one member of $\{O_{\beta}\}$. since this is open it must contain some $V_x$ with $x \in V_x$. now take the corresponding $O_x = \psi(V_x)$
the collection $\{O_x\}_{x \in X}$ is countable and covers $X$
A: What you are asking is to prove that every second countable space is Lindelof (in more common notation).
So, let's prove $\text{Second Countable}\implies\text{Lindelof}$. 
Let $X$ be second countable with countable basis $\mathscr{B}$, and let $\Omega=\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an open cover for $X$. By assumption, for each $\alpha\in\mathcal{A}$ we can cover $U_\alpha$ with some collection $\mathscr{B}_\alpha$ of elements of $\mathscr{B}$. Note then that $\displaystyle \Sigma=\bigcup_{\alpha\in\mathcal{A}}\mathscr{B}_\alpha$ is a countable open cover for $X$. So, for each element $O$ of $\Sigma$ choose an element $U$ of $\Omega$ containing it (by the axiom of choice) and define the function $f: \Sigma\rightarrow\{U_{\alpha}\}_{\alpha\in\mathcal{A}}$ by $f(O) = U$. Then, $\Gamma = f(\Sigma)$ is an open cover of $X$ (since its union contains the union over all the elements of $\Sigma$ which is $X$) and is countable since $f$ is a surjection and $\Sigma$ is countable. Thus, $\Gamma$ is our desired countable subcover of $\Omega$.
A: Let $\{V_1,V_2,\ldots\}$ be a countable open base of $M$, and let $\mathcal{U}=\{U_i\}_{i\in I}$ be an open cover of $M$. For each $n$, choose a set from $\mathcal{U}$ that contains $V_n$, call it $U_{i_n}$. If for some $n$ we cannot find a set from $\mathcal{U}$ that contains $V_n$ then so be it. Then, we will have some list:
$$V_1\subseteq U_{i_1}, \quad V_2\subseteq U_{i_2}, \quad V_3\subseteq X, \quad V_4\subseteq U_{i_4}, \quad \ldots$$
where $V_n\subseteq X$ represents the inability to find a set from $\mathcal{U}$ that contains $V_n$. Let $N\subseteq \mathbb{N}$ be the set of all $n$ for which $U_{i_n}$ exists.
We propose that the collection $\{U_{i_n}\}_{n\in N}$ is a countable subcover of $\mathcal{U}$. That it is countable is obvious, so we just need to show that $M\subseteq \bigcup_{n\in N} U_{i_n}$. Let $x\in M$ be given, then because $\mathcal{U}$ is an open cover of $M$, $x\in U_i$ for some $i\in I$. Moreover, since $U_i$ is open, it can be written as some union of $V_n$'s, i.e, $x\in V_n$ for some $n\in \mathbb{N}$. It must be that $n\in N$, otherwise, there is no set from $\mathcal{U}$ that contains $V_n$, this is a contradiction because $V_n\subseteq U_i$. Since $n\in N$, $U_{i_n}$ exists, therefore, $x\in V_n\subseteq U_{i_n}$, so $x\in \bigcup_{n\in N} U_{i_n}$ as desired. $\Box$
A: This is mainly inspired from David Holden's answer. I have mainly just organised it a bit more.
Also, by countable, I shall mean finite or countably infinite.

Let $\{O_\alpha\}_{\alpha \in \cal A}$ be the open cover that you had mentioned.
For each $n \in \mathbb{N},$ define
\begin{equation*} 
\mathcal{O}_n = \{O_\alpha : \alpha \in \mathcal{A}, V_n \subset O_\alpha\}.
\end{equation*}
(That is, the collection of all those elements of the original cover which contain $V_n$.)
Note that $\mathcal{O}_n$ need not be nonempty for every $n \in \mathbb{N}.$ (So far, we don't even claim that $\mathcal{O}_n$ is nonempty for any $n$ but our arguments later will show that it is. (Assuming $X \neq \varnothing$, of course.))
Let $N = \{n \in \mathbb{N} : \mathcal{O}_n \neq \emptyset\}.$
For any $n \in N,$ $\mathcal{O}_n$ is nonempty and thus, we can choose an element $U_n$ from it. (This is essentially using (some sort of) the axiom of choice.)
Observe that each $U_n$ is an element of $\{O_\alpha\}_{\alpha \in \mathcal{A}}.$ (That is, each $U_n = O_{\alpha_n}$ for some $\alpha_n \in \mathcal{A}.$)
Consider the countable subcover (it is a subcover by the above observation) given by
\begin{equation*} 
    \{U_n\}_{n \in N}.
  \end{equation*}
That this is countable is clear. We now show that it is a cover. (Note that we haven't claimed so far that $N \neq \emptyset.$)
Let $x \in X.$ We show that $x \in U_n$ for some $n \in N.$
Since $\{O_\alpha\}_{\alpha \in \mathcal{A}}$ is a cover of $X,$ there exists $\alpha \in \mathcal{A}$ such that $x \in O_{\alpha}.$
Since $\{V_n\}$ is a base, there exists $n \in \mathbb{N}$ such that $x \in V_n \subset O_\alpha.$
Clearly, $n \in N$ (since $O_\alpha \in \mathcal{O}_n$).  Moreover, $U_n \in \mathcal{O}_n$ and thus, $V_n \subset U_n.$
Since $x \in V_n,$ we see that $x \in U_n$ and hence, we see that $\{U_n\}_{n\in N}$ is a cover, as desired.
