I was looking up the proof that a closed subset of a compact space is itself a compact space (with the induced topology).
I came across this one:
Link
I believe I have found one step that might be something of an inaccuracy. How do we know that the open sets that comprise the open cover of $C$ are open in the ambient space $T$?
An example of what I mean: Say $T=[0,2]$ with the standard topology inherited from $\mathbb{R}$. And $C\subset T$ is given by $C=[0,1]$. Then $C$ is closed in $T$. Now $C$ is an open cover of itself but $C$ is not open in $T$. According to the proof $C \cup (T\setminus C) = [0,1] \cup (1,2] $ is then an open cover of $T$, but this is clearly not the case.
All help would be highly appreciated.
Best Answer
It really doesn't matter whether the covering subsets are open in $T$ (the ambient space) when we consider open covers of $C$. But if you insist on using subspace open sets:
let $\{U_i: i \in I\}$ be an arbitary open cover of $C$ (closed in the compact space $T$) using subsets that are open in $C$. By definition this means we can write each $U_i$ as $O_i \cap C$, where $O_i$ is open in $T$.
Now, $\{O_i: i \in I\} \cup \{T\setminus C\}$ is an open cover of $T$ by open sets of $T$. The latter holds, as $C$ is closed in $T$ (this is where this fact is crucially used). That it is a cover is clear: any point not in $C$ is covered by $T\setminus C$ and any point in $C$ is in some $U_i$ and so in the corresponding $O_i$ as well.
Compactness of $T$ tells us there is a finite subcover of that last cover, WLOG it contains $T \setminus C$ and is of the form $\{O_{i_1},\ldots O_{i_n}\} \cup \{T \setminus C\}$, for some finitely many $i_1,\ldots i_n$. It is then clear that $\{U_{i_1}, \ldots, U_{i_n}\}$ is a finite subcover of the original cover: if $x \in C$, it's certainly not covered by $T \setminus C$, so it must be in some $O_{i_j}$ ($j \in \{1,\ldots,n\}$) and thus also in $C \cap O_{i_j}=U_{i_j}$.
So an arbitary open cover of $C$ by $C$-open sets has a finite subcover, showing the compactness of $C$ in its subspace topology.