Show A is locally compact, if A is a closed subset of a locally compact space. Given X a locally compact space and $A \subset X$ a closed subset. How do I show that A is locally compact as well? I have no clue whatsoever on how I should handle this!

Let $x\in A$, a there exists a compact neighborhood $C\subset X$ of $X$ of $x$ in $X$. $C\cap A$ is a compact neighborhood of $x$ in $A$. We deduce that $A$ is locally compact.

To show that $C\cap A$ is compact, consider a covering $(U_i)_{i\in I}$ of $C\cap A$. You can suppose that there exists an open subset $V_i$ of $X$ such that $U_i=V_i\cap A$, Since $A$ is closed, $A\cap C$ is closed and $W=X-A\cap C$ is open. This implies that $W\bigcup \bigcup_{i\in I}V_i$ is an open covering of $C$. You can extract a finite covering $W, V_1,...,V_n$. This implies that $U_1,..U_n$ cover $A\cap C$.