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?
Let $(U_i)_{i\in I}$ be an open cover of $A$, $U_i=V_i\cap A$ where $V_i$ is open in $X$, $(V_i)_{i\in I}\cup (X-A)$ is an open cover of $X$ and has a finite subcover $V_1,...,V_n, X-C$, $A=X\cap A)=(V_1\cup..\cup V_n\cup (X-A))\cap A= U_1\cup..\cup U_n$.