Proof that the continuous image of a connected set is connected

Proof

We argue by contradiction, and pull a splitting of $f(S)$ back to $S$.

1. Recall that a subset is connected when it is a connected space in the subspace topology. The open sets of $f(S)$ are the sets $f(S) \cap U$ with $U$ open in $Y$.
2. Suppose that $f(S)$ is not connected. Then there are open $U, V \subseteq Y$ such that $U' = f(S) \cap U$ and $V' = f(S) \cap V$ are nonempty and disjoint, with $U' \cup V' = f(S)$.
3. Let $P = S \cap f^{-1}(U)$ and $Q = S \cap f^{-1}(V)$, with preimages taken under $f$. Since $f$ is continuous, $f^{-1}(U)$ and $f^{-1}(V)$ are open in $X$. So $P$ and $Q$ are open in $S$.
4. $P$ is nonempty. Take $y \in U'$. Then $y = f(s)$ for some $s \in S$, and $s \in P$. In the same way, $Q$ is nonempty.
5. $P$ and $Q$ are disjoint. If $s \in P \cap Q$, then $f(s) \in f(S) \cap U \cap V = U' \cap V'$, which is empty.
6. $P \cup Q = S$. For $s \in S$, $f(s) \in f(S) = U' \cup V'$, so $f(s) \in U$ or $f(s) \in V$. Thus $s \in P$ or $s \in Q$.
7. So $P$ and $Q$ split $S$, and $S$ is not connected. This contradicts the hypothesis, so $f(S)$ is connected.