Show that if $X$ is compact, and if $A$ is a closed subset of $X,$ then $A$(with the subspace topology ) is also compact.

**A HINT:**

Use the subspace topology to relate open sets in $A$ with open sets in $X;$ somehow you need to find an open cover of $X.$

**Still I am unable to solve it using the hints given, could anyone help me in solving it using the hints given?**