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

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> 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.
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Now the proof

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> 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
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> $$(X\setminus A) \cup \left(\bigcup\_{j \in J} U\_j\right)=X$$
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> Then $A \subseteq \bigcup\_{j \in J} U\_j$ and by the lemma, we have $A$ to be compact.
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Now, I think I can shorten this as follows

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> 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.
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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?