MHB Why is $X$ connected if $A$ and $X/A$ are connected in a topological group?

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Here is this week's POTW:

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Let $X$ be a topological group; let $A$ be a subgroup of $X$ such that $A$ and $X/A$ are connected. Show that $X$ is connected.

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No one answered this week's problem. You can read my solution below.

Spoiler

Suppose $\{U,V\}$ is a separation of $X$. The natural projection map $p: X \to X/A$ is open, so since $p(U) \cup p(V) = X/A$ and $X/A$ is connected, then $p(U) \cap p(V)$ is nonempty. That means there exists $x\in X$, $u\in U$, and $v\in V$ such that $u,v\in xA$. Then $\{U\cap xA, V\cap xA\}$ separates $xA$, which is impossible since $xA$ is connected (as $A$ is connected and left translation by $x$ is a homeomorphism of $X$).