Every closed subspace of a compact space is compact Let $(X,\tau_{X})$ be a compact topological space and let $A \subset X$, $\tau_{X}$ - closed. We need show that $(A, \tau_{A})$ is a compact topological subspace.
Proof:
Let $U = \{U_\lambda\}_{\lambda \in L} $ be a open cover of $X$. Then, there exists $\{U_1, \ldots, U_k\} \subset  U$ such that:
X = $U_1 \cup \cdots \cup U_k \implies  A = X \cap A = ( U_1 \cup ... \cup U_k) \cap A = \cup V_\lambda ; V_\lambda = U_\lambda \cap A \in \tau_{A}.$ Therefore, (A, $\tau_{A}$) is a compact  subspace.
In particular, $U\cap A$ is a $\tau_{A}$ - open conver of $A.$
Is it correct?
 A: What you want to show is that for any open cover $\bigcup_{i \in J} V_j $ of $A$ you can find a finite subcover $A \subset \bigcup_{i=1}^n V_i$. So to answer your question no it is not entirely right. Because
what you proved is that if you have a finite subcover for $X$ you have one for $A$, which is only a part of the whole proof which should go as:
Since $A$ is closed $X\backslash A$ is open, and thus $\{ X \backslash A , \bigcup_{i \in J} V_{i}\}$ is an open cover of $X$. Compactness of $X$ allows you to find a finite subcover $\{ X \backslash A , \bigcup_{i=1}^n V_{i}\}$ $\textbf{denoted that the index goes now up to $n$ thus finite}$. Because $A \subset X$  we have that $A \subset \bigcup_{i=1}^n V_{i}$ and thats what we wanted. $\{V_{i}\}^n_{i=1}$ is a finite subcover of A and $V_i$ was arbitrary thus $A$ is compact.
A: No, that's incorrect. You need to start from an open cover of $A$, not of $X$.
Suppose $\mathcal{V}$ is an open cover of $A$, so the sets $V\in\mathcal{V}$ are open in $X$ and
$$
A\subseteq \bigcup \mathcal{V}
$$
Now consider $\mathcal{V}'=\mathcal{V}\cup(X\setminus A)$, which is obviously an open cover of $X$ and conclude: you can find $V_1,\dots,V_n\in\mathcal{V}'$ such that
$$
X=V_1\cup V_2\cup\dots\cup V_n
$$
If $X\setminus A$ is not among the $V_i$, you're done. Otherwise, without loss of generality $X\setminus A=V_n$ and
$$
A\subseteq V_1\cup\dots\cup V_{n-1}
$$
