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  "body": "\n## Proof\n\nWe argue by contradiction, and pull a splitting of $f(S)$ back to $S$.\n\n1. Recall that a subset is [[connected|0R37DFJKFW]] when it is a [[connected space|0D939CMME7]] in the [[subspace topology|0D5YHQGDJY]]. The open sets of $f(S)$ are the sets $f(S) \\cap U$ with $U$ open in $Y$.\n2. 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 $U' \\cup V' = f(S)$.\n3. Let $P = S \\cap f^{-1}(U)$ and $Q = S \\cap f^{-1}(V)$, with [[preimages|0FGCFY1AZX]] taken under $f$. Since $f$ is [[continuous|0SY9ECQ19Q]], $f^{-1}(U)$ and $f^{-1}(V)$ are open in $X$. So $P$ and $Q$ are open in $S$.\n4. $P$ 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.\n5. $P$ 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.\n6. $P \\cup Q = S$. 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 $s \\in Q$.\n7. So $P$ and $Q$ split $S$, and $S$ is not connected. This contradicts the hypothesis, so $f(S)$ is connected.\n\n## Idea\n\nContinuity means that preimages of open sets are open. So a splitting of the image into two open pieces pulls back to a splitting of $S$, which cannot exist.\n",
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