1. **Problem statement:** Show that if $X$ is Hausdorff, then every retract $A$ of $X$ is closed in $X$. 2. **Definitions and formula:** A retraction $\phi: X \to A$ is a continuous map such that $\phi(a) = a$ for all $a \in A$. 3. **Step 1: Use the property of Hausdorff spaces.** In a Hausdorff space, points can be separated by neighborhoods, and singletons are closed. 4. **Step 2: Show $A$ is closed.** Consider the map $\phi: X \to A$. Since $\phi$ is continuous and $A$ is the image of $\phi$, we have $$A = \{x \in X : \phi(x) = x\} = \{x \in X : x - \phi(x) = 0\}.$$ 5. **Step 3: Use the diagonal argument.** Define the map $f: X \to X \times X$ by $f(x) = (x, \phi(x))$. The diagonal $\Delta = \{(a,a) : a \in A\}$ is closed in $X \times X$ because $X$ is Hausdorff. 6. **Step 4: Express $A$ as a preimage.** Then $$A = f^{-1}(\Delta)$$ which is the preimage of a closed set under a continuous map, so $A$ is closed in $X$. **Final answer:** If $X$ is Hausdorff, every retract $A$ of $X$ is closed in $X$.