I am baffled with what I am seeing. First, here's what is noted as a definition in my notes
Let $X$ be a set and $A \subseteq X$. A cover of $A$ by subsets of $X$ is a family $(W_i)_{i \in I}$ of subsets of $X$ such that $A \subseteq \cup_{i \in I} W_i$. For $J \subseteq I$ we say that $(W_j)_{j \in J}$ is a subcover of $(W_i)$ if itself is a cover of $A$ by subsets of $X$.
Sure. Got it. However, the following which to me, seems like a definition is given as a lemma
$A \subseteq X$ is compact if and only if every cover of $A$ by open subsets of $X$ has a finite subcover.
That's essentialy the definition of "compactness" .... The proof (forward) I looked at and it still is strange
Suppose $A$ is compact. Let $(W_i)_{i \in I}$ be a cover of $A$ by open subsets of $X$. For each $i \in I$, put $U_i = A \cap W_i$. BY the suspace topology, $U_i$ is open in $A$. Then,
$$A=(\cup W_i) \cap A=\cup(W_i \cap A)=\cup U_i$$
Hence $(U_i)$ is an open cover of $A$. Since $A$ is compact, we can choose finite $J \subseteq I$ such that $A=\cup U_j$. But $U_j \subseteq W_j$ for all $j$ so $A \subseteq \cup W_j$
Well, why go through the trouble? I mean it's said that $A$ is compact by assumption i.e. every open cover of $A$ has a finite subcover so if there is an open cover $(W_i)$, then there is a finite subcover for sure.
I mean what is going on with this bizarre proof of a definition? A compactness of a subspace is a lemma? Not given by a definition? I am so confused, then why is a the general definition of a compact space $X$ given without proof? i.e. Why is it not
$X$ is compact if and only if every open cover has a finite subcover and now here's the proof...
As much as the above doesn't make sense since that's how we acknowledge the notion, the proof for $A$ doesn't make sense to me.