Prove all closed subspace of a compact space are compact: Redundancy?

Question

I see a redundancy in the following proof of the statement. First, we have a lemma that this proof uses

A subspace $A \subseteq X$ is compact if and only if every open cover of $A$ by open subsets of $X$ has a finite subcover.

Now the proof

Let $X$ be compact and $A \subseteq X$ be closed. Let $(U_i)_{i \in I}$ be a cover of $A$ by open subsets of $X$. Then $(U_i)$ together with $X \setminus A$ is an open cover of $X$. Since $X$ is compact, it has some finite subcover, $J \subseteq I$ that is finite such that

$$(X\setminus A) \cup \left(\bigcup_{j \in J} U_j\right)=X$$

Then $A \subseteq \bigcup_{j \in J} U_j$ and by the lemma, we have $A$ to be compact.

Now, I think I can shorten this as follows

Let $X$ be compact and $A \subseteq X$ be closed. Let $(U_i)_{i \in I}$ be a cover of $A$ by open subsets of $X$. Since $X$ is compact, there exists some finite $J \subseteq I$ such that $\bigcup U_j=X$. Since $A \subseteq X$, we clearly have $A \subseteq \bigcup_{j \in J} U_j$ as required.

Well..? Why do we need to consider the complement $X\setminus A$? Why put that in the finite open cover that we already have? Shouldn't $\bigcup U_j$ suffice on its own as a finite open cover of $A$ anyway? I do't see the role of $X\setminus A$ at all in the proof. can someone explain?

Answer 1

Since $X$ is compact, there exists some finite $J\subseteq I$ such that $\bigcup U_j=X$.

That is the mistake: The open cover that covers $A$ was not assumed to cover all of $X$, so it's not an open cover of $X$. If it doesn't cover $X$ then it can't have a finite subset that covers $X$.

Answer 2

There's no reason why you would have $\bigcup_i U_i = X$. usually it's false.

In your argument, you don't use the fact that $A$ is closed, so if it was correct, it would work with any $A$. And $]0,1[ $ is not compact while $[0,1]$ is