Let $X$ be a Hausdorff space and $A \subset X$ compact. Show that $A$ is closed.
Pick $x \in X \setminus A$, then since $X$ is Hausdorff there exists disjoint $O_x$ and $O_A$. Since $O_x \subset X \setminus A \implies X \setminus A$ is open. Thus $A$ is closed.
I'm not following how can they make the assumption that $O_x \subset X \setminus A$? Even though $x \in X \setminus A$ I don't think that implies that the neighbourhood of $X$ must be in $X \setminus A$?
@Lakri tsa Hausdorff space means any two points are "strongly" separated by two open sets. You can't separate a point from a set. But since $T_2\subset T_3$ , you can separate a point from a closed set not containing that point. And you can't assume $A$ is closed upfront.
$A\subset X$ compact and $X$ Hausdorff implies $A\subset X$ is closed.
Hint: $x\in X\setminus A$ .
Then, for any point $a\in A $ $\exists U_{x_a}, U_a $ such that $U_{x_a} \cap U_a=\emptyset$
$\{U_a:a\in A\}$ is a open cover of $A$.
Suppose, $\{U_{a_i}:i\in I \}$ is a finite subcover.
Choose, $U=\cap_{i\in I}U_{x_{a_i}}$
Then, $U \subset X\setminus A$