Continuous Image of Connected Space is Connected

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.

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

Suppose, for the sake of contradiction, that $S_2=f\left(S_1\right)$ is disconnected. Then 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 $T_1$. Suppose $x \in U_1 \cap V_1 \cap S_1$. Then $f(x) \in U_2\cap V_2 \cap S_2$, a contradiction. Thus $S_1$ is disconnected, contradicting the hypothesis.

Another proof
Let $i: f \left({T_1}\right) \to T_2$ be the inclusion mapping.

Let $g: T_1 \to f \left({T_1}\right)$ be the surjective restriction of $f$.

Then $f = i \circ g$.

Hence, by Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $g$ is continuous.

We use a proof by contradiction.

Suppose that $A \mid B$ is a partition of $f \left({T_1}\right)$.

Then it follows that $g^{-1} \left({A}\right) \mid g^{-1} \left({B}\right)$ is a partition of $T_1$.