I was looking at the following proposition :
"Every closed subspace of a compact topological space is compact"
and I am wondering why the following proof is not good :
Let $(X, \tau)$ be a compact topological space and $C \subset X$ be a closed set in $(X, \tau)$ endowed with the induced topology $\tau_{C}$.
Let $A$ be an arbitrary set. For any open cover $\mathcal{U} = (U_{\alpha})_{\alpha \in A}$ of $(X, \tau)$, $\mathcal{V} = (C \cap U_{\alpha})_{\alpha \in A} = (V_{\alpha})_{\alpha \in A}$ is an open cover of $(C, \tau_{C})$.
Since $(X, \tau)$ is compact, for any open cover $(U_{\alpha})_{\alpha \in A}$ of $(X, \tau)$, there exists a finite subcover $(U_{\alpha_{i}})_{i = 1, ..., n}$.
So $(V_{\alpha_{i}})_{i = 1, ..., n}$ is a finite subcover for $(C, \tau_{C})$. $\square$
In that proof, we don't need the fact that $C$ is closed, but I cannot manage to convince myself that it's wrong (because if I replace $C$ by an open set, I think it works also and it is obviously a false statment). No need to give me the usual proof (https://proofwiki.org/wiki/Closed_Subspace_of_Compact_Space_is_Compact) that I understand, I just don't get why the one above is not good.
It is not a good proof because you didn't prove that any open cover of $C$ has a finite subcover. You showed that if you take an open cover of $X$ and intersect all the sets in it with $C$ then you have a finite subcover. But what if you build an open cover of $C$ in a different way? For example, if you take the set $(0,1)$ then the open cover $\{(\frac{1}{n},1)\}_{n=1}^\infty$ has no finite subcover, despite $(0,1)$ being a subspace of $[0,1]$.