Suppose that for every $n\in\mathbb N$, $V_n$ is a non-empty compact subset of a Hausdorff space $X$ and suppose that $V_{n+1} \subset V_n$, then $\bigcap V_n$ is nonempty.
The proof goes as follows:
Proof: suppose that the intersection is empty, $\bigcap V_n = \emptyset$.
Now let $U_n= V_1∖V_n$. The $\{U_n\}$ forms a covering of $V_1$ since $\bigcap V_n = \emptyset$.
The conclusion is that since that since $V_1$ is compact there exists a finite subcover of these $U_i$'s.
But, the $U_i$'s are not necessarily open in $X$. They are open in the subspace topology of $V_1$. However, $V_1$ is compact in $X$ but not necessarily itself.
How are we able to pull out the finite subcover?
For $U_i$ to be open in the subspace topology of $V_1$, there must be some $\hat{U}_i$ which is open in $X$ and has $\hat{U}_i\cap V_1=U_i$ (this is how the subspace topology is defined). Then $\{\hat{U}_i\}$ forms an open cover of $V_i$ in the sense of $X$.
Put another way, if $A$ is a subspace of $B$ then $A$ is compact in itself with the subspace topology iff $A$ is compact in $B$.