Continuous Image of Connected Space is Connected/Proof 2

Theorem
Let $T_1$ and $T_2$ be topological spaces, and let $S_1 \subseteq T_1$ be connected.

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

Then the image $f \left({S_1}\right)$ is connected.

Proof
By hypothesis, $f: T_1 \to T_2$ is continuous.

Suppose that $S_2=f\left(S_1\right)$ is disconnected.

Then by Definition:Connected (Topology)/Subset/Definition 2 there are open sets $U_2$ and $V_2$ in $T_2$ such that $S_2 \in U_2 \cup V_2$, $U_2 \cap V_2 \cap S_2 = \varnothing$, $U_2 \cap S_2 \ne \varnothing$, and $V_2 \cap S_2 \ne \varnothing$. Then $U_1=f^{-1} \left({U_2}\right)$ and $V_1=f^{-1} \left({V_2}\right)$ are open in $T_1$.

$U_2 \cap S_2 \ne \varnothing$, so for some $x \in S_1$, $f(x) \in U_2$. Then $x \in f^{-1} \left({U_2}\right) = U_1$ and $x \in S_1$, so $U_1 \cap S_1 \ne \varnothing$.

Similarly, $V_1 \cap S_1 \ne \varnothing$.

If for some $x \in S_1$, $x \in U_1 \cap V_1 \cap S_1$, then $f(x) \in U_2\cap V_2 \cap S_2$, a contradiction, so $U_1 \cap V_1 \cap S_1 = \varnothing$.

Thus $S_1$ is disconnected.