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  "body": "\n## Proof\n\nWe argue by contradiction, and restrict a splitting of $X$ to the image of a path.\n\n1. Suppose that $X$ is not [[connected|0D939CMME7]]. Then there are open $U, V \\subseteq X$, both nonempty and disjoint, with $U \\cup V = X$.\n2. Choose $x \\in U$ and $y \\in V$. Since $X$ is [[path-connected|0K0GE6BCYW]], there is a continuous $\\gamma \\colon [0, 1] \\to X$ with $\\gamma(0) = x$ and $\\gamma(1) = y$.\n3. The set $[0, 1]$ is an [[interval|0TK8Y3J5RW]], so it is a [[connected subset|0R37DFJKFW]] of $\\mathbb{R}$, by [[connected subsets of the real line are intervals|0B9ZQFMQNV]]. So $[0, 1]$ with its [[subspace topology|0D5YHQGDJY]] is a connected space.\n4. By [[continuous image of a connected set is connected|0EPD87YVG8]], applied to $\\gamma$ on the space $[0, 1]$, the set $K = \\gamma([0, 1])$ is connected.\n5. The sets $K \\cap U$ and $K \\cap V$ are open in $K$. They are disjoint, because $U \\cap V = \\varnothing$, and their union is $K$, because $U \\cup V = X$.\n6. They are nonempty, because $x = \\gamma(0) \\in K \\cap U$ and $y = \\gamma(1) \\in K \\cap V$.\n7. So $K$ is not connected, which contradicts step 4. Thus $X$ is connected.\n\n## Idea\n\nA path is a continuous image of the connected interval $[0, 1]$, so its image is connected. If $X$ split into two open pieces, the path from a point of one piece to a point of the other would split too.\n",
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