We must first define the notion of covers
Let X be a set, and \mathscr{A} be a collection of sets. Then \mathscr{A} is called a cover of X if
X \subseteq \bigcup \mathscr{A}
Let X be a set and \mathscr{A} be a cover of X. If there exists a collection \mathscr{B} \subseteq \mathscr{A} that also covers X, then \mathscr{B} is a subcover of \mathscr{A} for X.
These definitions are true of any set. Next we show how we will use these results in topology.
Let S = (X,\mathcal{T}) be a topological space, and \mathscr{A} be a cover of X. Then we call \mathscr{A} an open cover if
\mathscr{A} \subset \mathcal{T}.
Thus, \mathscr{A} is an open cover if it is made up of open sets.
Now we define the notion of a space being compact.
A space X = (S,\mathcal{T}) is compact if every open cover \mathscr{A} of S has a finite subcover that also covers S.
We next prove that subspaces are compact by the same definition.
Let X be a space and Y be a subspace of X. Then Y is compact if and only if every cover of Y by open sets in X has a finite subcover.
(\leftarrow) Suppose Y is compact, and \mathscr{A} = \{A_\alpha\}_{\alpha \in I} is a cover of Y by sets open in X. Then the collection
\{A_\alpha \cap Y \mid \alpha \in I\}
is a cover of Y by sets open in Y, making it an open cover of Y. Since we assumed Y is compact, we have that there exists a finite subcover
\{A_{\alpha_1} \cap Y, \dots, A_{\alpha_n} \cap Y\}.
Then \{A_{\alpha_1}, \dots, A_{\alpha_n}\} is a finite subcover of Y by sets open in X.
(\rightarrow) Suppose every cover of Y by open sets in X has a finite subcover. Let \mathscr{B} = \{B_{\alpha}\}_{\alpha \in I} be an open cover of Y. Then, for each \alpha \in J, choose a set A_\alpha open in X such that
A_\alpha = B_\alpha \cap Y.
(To-DO: Prove such a set exists) Then the collection \mathscr{A} = \{A_\alpha\}_{\alpha \in I}. is a cover of Y by sets open in X. Thus there exists a finite subcover \{A_{\alpha_1}, \dots, A_{\alpha_n}\}. Then \{B_{\alpha_1}, \dots, B_{\alpha_n}\} is a finite subcover of Y, and Y is compact.
Next we prove many properties of compact spaces
Every closed subspace of a compact space is compact.
Let X be a compact topological space and Y be a closed subspace of X. Then we have that Y is closed in X, implying X \backslash Y is open in X by definition.
Let \mathscr{A} be a cover of Y by sets open in X. Then
\mathscr{B} = \mathscr{A} \cup X \backslash Y
is an open cover of X. Since X is compact, there exists a finite subcover \mathscr{B}^\prime which covers Y. We may safely remove X\backslash Y from \mathscr{B}^\prime to obtain a finite subcover of \mathscr{A}. Thus Y is compact.