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  "body": "\n## Proof\n\nWe combine the two facts that images of connected sets are connected and that connected subsets of $\\mathbb{R}$ are intervals.\n\n1. The subspace topology on $X$ as a subset of itself is the topology of $X$. So $X$ is a [[connected subset|0R37DFJKFW]] of $X$, because $X$ is a [[connected space|0D939CMME7]].\n2. The map $f$ is [[continuous|0SY9ECQ19Q]]. By [[continuous image of a connected set is connected|0EPD87YVG8]], the set $f(X) \\subseteq \\mathbb{R}$ is connected.\n3. By [[connected subsets of the real line are intervals|0B9ZQFMQNV]], $f(X)$ is an interval. So for $p, q \\in f(X)$ and $p \u003c c \u003c q$, we have $c \\in f(X)$.\n4. Let $c$ be between $f(a)$ and $f(b)$. If $c = f(a)$, take $x = a$. If $c = f(b)$, take $x = b$.\n5. Otherwise $c$ lies strictly between the two values. Let $p$ be the smaller and $q$ the larger of $f(a)$ and $f(b)$. Then $p, q \\in f(X)$ and $p \u003c c \u003c q$.\n6. By step 3, $c \\in f(X)$. So there is $x \\in X$ with $f(x) = c$.\n\n## Idea\n\nThe image of a connected space under a continuous map is connected. In $\\mathbb{R}$ the connected sets are the intervals, and an interval that contains $f(a)$ and $f(b)$ contains each value between them.\n",
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