I am reading a proof of a proposition about compactification.
In the proof, I guess the author uses the following proposition.
Proposition 1:
Let $S$ be a topological space.
Let $C_\lambda$ be compact closed subspaces of $S$ for any $\lambda\in\Lambda_1$.
Let $A_\lambda$ be closed subspaces of $S$ for any $\lambda\in\Lambda_2$.
Then, $(\bigcap_{\lambda\in\Lambda_1} C_\lambda)\,\bigcap\,(\bigcap_{\lambda\in\Lambda_2} A_\lambda)$ is a compact closed subspace of $S$.
I think the above proposition is true.
I proved the above proposition as follows.
Is my proof ok or not?
First, I prove the following proposition:
Proposition 2:
Let $S$ be a topological space.
Let $C$ be a compact subspace of $S$.
Let $A$ be a closed subspace of $S$.
Then, $C\cap A$ is a compact closed subspace of $S$.
My proof of Proposition 2:
I use the fact that a closed subspace of a compact space is compact.
Since $A$ is closed in $S$, $C\cap A$ is closed in $C$.
Since $C\cap A$ is closed in a compact space $C$, $C\cap A$ is a compact in $C$.
So, $C\cap A$ is a compact closed subspace of $C$.
So, $C\cap A$ is a compact closed subspace of $S$.
My proof of Proposition 1:
$(\bigcap_{\lambda\in\Lambda_1} C_\lambda)\,\bigcap\,(\bigcap_{\lambda\in\Lambda_2} A_\lambda)=C_{\lambda_0}\cap (\bigcap_{\lambda\in\Lambda_1-\{\lambda_0\}} C_\lambda)\,\bigcap\,(\bigcap_{\lambda\in\Lambda_2} A_\lambda)$.
$(\bigcap_{\lambda\in\Lambda_1-\{\lambda_0\}} C_\lambda)\,\bigcap\,(\bigcap_{\lambda\in\Lambda_2} A_\lambda)$ is a closed subspace of $S$.
And $C_{\lambda_0}$ is a compact subspace of $S$.
By Proposition 2, $(\bigcap_{\lambda\in\Lambda_1} C_\lambda)\,\bigcap\,(\bigcap_{\lambda\in\Lambda_2} A_\lambda)$ is a compact closed subspace of $S$.
Your proofs are correct, in both cases you are using by definition of subspace topology that if $A\subset S$ is a closed(open) subspace of S , given $C$ compact, $C\cap A$ is a closed set in S, and compactness can be checked on a open cover(of open set in the subspace) of a subspace. Moreover you are using an hypothesis valid always which is that closed in a compact is compact.