Continuous Image of Separable Space is Separable

Definition
Let $T_1 = \left({S_1, \tau_1}\right), T_2 = \left({S_2, \tau_2}\right)$ be topological spaces.

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

If $T_1$ is separable, then so is the image $f \left[{T_1}\right]$.

That is, separability is a continuous invariant.

Proof
By definition, $T_1 = \left({S_1, \tau_1}\right)$ is separable there exists a countable subset $D \subset S_1$ which is (everywhere) dense in $T_1$.

We need to show that if there exists a continuous mapping $f: T_1 \to T_2$, then $f \left[{T_1}\right]$ is also separable.

That is, that there exists a countable subset of $f \left[{S_1}\right]$ which is dense in $T_2$.

Let $x_2$ be any point in the image $f \left[{S_1}\right]$ of $S_1$ under $f$.

Let $U \in \tau_2$ be an arbitrary open set such that $x_2 \in U$.

By definition of image set, there exists some $x_1 \in S_1$ with $f \left({x_1}\right) = x_2$.

Since $f$ is continuous, $f^{-1} \left[{U}\right]$ is open in $T_1$.

By definition of preimage, $x_1$ is in this set.

We are given that $D$ is a countable subset of $S_1$ which is dense in $T_1$.

By definition of dense, $D \cap f^{-1} \left[{U}\right] \ne \varnothing$

Thus there exists some $d \in D$ such that $d \in f^{-1} \left[{U}\right]$.

Therefore $f \left({d}\right) \in U$.

Since $U$ was arbitrary, it follows that $f \left[{D}\right]$ is dense in $T_2$.

By Image of Countable Set under Mapping is Countable, $f \left[{D}\right]$ is countable.

Hence $T_2$ is separable.