Continuous Image of Separable Space is Separable

Definition
Let $T_1 = \left({X_1, \tau_1}\right), T_2 = \left({X_2, \tau_2}\right)$ be topological spaces.

Let $f: T_1 \to T_2$ be a continuous surjective mapping.

If $T_1$ is separable, then so is $T_2$.

Proof
From the definition of separable, $T_1 = \left({X_1, \tau_1}\right)$ is separable if there exists a countable subset $D \subset X_1$ which is everywhere dense.

We need to show that if there exists a mapping $f: T_1 \to T_2$ which is continuous, then $T_2$ is also separable.

That is, there exists a countable subset of $X_2$ which is everywhere dense.

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=== Edit, Tom Collinge 17 Feb 2016 ======================

Let $x_2 \in X_2$ be any point in $X_2$, and $O$ with $x_2 \in O \in \mathscr \tau_2$ be any open set containing $x_2$.

Since $f$ is surjective there is (at least one) $x_1 \in X_1$ with $f(x_1) = x_2$

Since $f$ is continuous, $f^{-1}(O)$ is an open set $\in \mathscr \tau_1 $ with $x_1 \in f^{-1}(O)$.

Since $D$ is dense in $X_1$ there is some $d \in D$ with $d \in f^{-1}(O)$.

Therefore $f(d) \in O$ and so $f(D)$ is dense in $X_2$, $f(D)$ is countable ($|f(D)| \le |D|)$, and so $T_2$ is separable.

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=== End Edit 17 Feb 2016 ================================