Second-Countability is Preserved under Open Continuous Surjection

Theorem
Let $T_A = \left({S_A, \tau_A}\right)$ and $T_B = \left({S_B, \tau_B}\right)$ be topological spaces.

Let $\phi: T_A \to T_B$ be a surjective open mapping which is also continuous.

If $T_A$ is second-countable, then $T_B$ is also second-countable.

Proof
Let $\phi$ be surjective, continuous and open.

Let $T_A$ be second-countable.

By definition of second-countability $T_A$ has a countable basis, $\mathcal B$, say.

Let $\mathcal B = \left\{{V_n: n \in \N}\right\}$.

We need to show that $\left\{{\phi \left[{V_n}\right]: n \in \N}\right\}$ is a base for $T_B$.

Let $U$ be an open set of $T_B$.

$\phi$ is continuous, so $\phi^{-1} \left[{U}\right]$ is open in $T_A$.

As $\mathcal B$ is a base for $T_A$, there exists an open set $V_n \subseteq \phi^{-1} \left[{U}\right]$.

$\phi$ is surjective, so from Surjection iff Right Inverse we have that:
 * $\phi \left({\phi^{-1} \left({U}\right)}\right) = U$

So, applying $\phi$ to $V_n$, from we obtain:
 * $\phi \left[{V_n}\right] \subseteq U$.

This means that $\left\{{\phi \left[{V_n}\right]: n \in \N}\right\}$ is a base for $T_B$.

Thus, $T_B$ is second-countable.