Continuous Image of Connected Space is Connected/Proof 2

Proof
Suppose that $S_2 = f \sqbrk {S_1}$ is not connected in $T_2$.

Then by definition there exist open sets $U_2$ and $V_2$ in $T_2$ such that:
 * $S_2 \subseteq U_2 \cup V_2$
 * $U_2 \cap V_2 \cap S_2 = \O$
 * $U_2 \cap S_2 \ne \O$
 * $V_2 \cap S_2 \ne \O$

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

Thus $U_1 = f^{-1} \sqbrk {U_2}$ and $V_1 = f^{-1} \sqbrk {V_2}$ are open in $T_1$.

We have that:
 * $U_2 \cap S_2 \ne \O$

Therefore:
 * $\exists x \in S_1: \map f x \in U_2$

Then:
 * $x \in f^{-1} \sqbrk {U_2} = U_1$

and:
 * $x \in S_1$

so:
 * $U_1 \cap S_1 \ne \O$

Similarly:
 * $V_1 \cap S_1 \ne \O$

Suppose there exists $x \in S_1$ such that $x \in U_1 \cap V_1 \cap S_1$.

Then:
 * $\map f x \in U_2 \cap V_2 \cap S_2$

which is a contradiction.

It follows that:
 * $U_1 \cap V_1 \cap S_1 = \O$

Thus by definition $S_1$ is not connected in $T_1$.

The result follows by the Rule of Transposition.