$U\cap K$ is compact in the subspace topology

Question

If $X,\tau$ is a topological space and K is a closed subspace. If $U$ is a compact set of $X,\tau$ then $U\cap K$ is compact.

Attempted proof:

Since $U$ is compact there is a finite sub covering such that $U\subset \bigcup_\limits{i=1}^{n}B_i$ $B_i\in \tau\forall 1\leqslant i\leqslant n$

$U\cap K\subset (\bigcup_\limits{i=1}^{n}B_i)\cap K=\bigcup_\limits{i=1}^{n}(B_i\cap K)$

$B_i\cap K$ is open for the subspace topology for all $i$.

Hence $\bigcup_\limits{i=1}^{n}(B_i\cap K)$ is a finite sub covering of $U\cap K$ proving $U\cap K$ to be compact in the subspace topology.

Question:

Is my proof right? If not how should I correct it?

Thanks in advance!

Answer 1

The proof is wrong from the start. Every topological space has a finite covering: just take the covering with a single open set, which is the whole space.

Let $(O_\lambda)_{\lambda\in\Lambda}$ be an open cover of $U\cap K$. Add to this cover the set $U\cap K^\complement$ (which is an open subset of $U$). Then you have an open cover of $U$. Since $U$ is compact, it has a finite subcover. The elements of this finite subcover which are distinct from $U\cap K^\complement$ form a finite subcover of $U\cap K$.