How can you prove that a retract of a Hausdorff space is always closed?

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Here is this week's POTW:

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Give two different proofs of the following result: Every retract of a Hausdorff space is closed.

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

Spoiler

Let $X$ be a Hausdorff space, $A\subset X$ a subspace, and $r : X \to A$ a retraction. I'll show $A$ is closed in two ways.

Proof 1. Since $X$ is Hausdorff, its diagonal $\Delta$ is closed. Since $r$ is continuous and $A = (\operatorname{id},r)^{-1}(\Delta)$, then $A$ is closed.

Proof 2. Let $x$ be a closure point of $A$. There is a net $\{x_\alpha\}$ in $A$ which converges to $x$. Continuity of $r$ implies $r(x_\alpha)$ converges to $r(x)$. Since $r$ is a retraction, $r(x_\alpha) = x_\alpha$. Hence, $\{x_\alpha\}$ converges to $r(x)$ as well. The Hausdorff property of $X$ implies $x = r(x)$. In particular, $x\in A$. As $x$ was arbitrary, $A$ is closed.