Compactness of subsets of a topological space I think I know the answer, but there is a warning below the question that it is not as trivial as seems.
My question is: verify:

Let $A$ be a subset of $(X, T) $ and $T_1$ the topology induced on $A$ by $T$. Then $A$ is a compact subset of $(X, T)$ if and only if $(A, T_1)$ is a compact space.

The definition I have for a compact subset is a subset of a topological space in which every open covering has a finite subcovering.
The definition I have for a compact space is a space $(X, T)$ in which a compact subset equals $X$.
My attempt at solving:
If $A$ is a compact subset of $(X, T)$, then every open covering of $A$ has a finite subcovering in $X$. Since $ T_1$ is the induced topology, every set in $A$ that was open in $(X, T)$ is open in $(A, T_1)$, as well as the intersection of any open subset of $(X, T)$ with $A$. Then $A$ has an open covering in $(A, T_1)$ where the union of the open sets in the covering equals $A$. So $A$ is a compact subset of $(A, T_1)$
If $(A, T_1)$ is a compact space, it contains a compact subset $B$ equal to $A$. Since the sets in the open coverings of $B$ in $(A, T_1)$ are subsets of sets that form a collection that covers $A$ in $(X, T)$ (because $T_1$ is the induced topology: the intersection of $A$ and another set in $X$ is a subset of $X$). Thus $A$ is a compact subset of $(X, T)$
Is there anything wrong with my assumptions/reasoning? Also any advice on writing a better proof? A better Stack Exchange question? I'm new to this, so please point out anything I could do better.
 A: The first and second point are a bit unclear to me. The first can be divided into three points:

*

*$A$ is compact in $(X,T)$, therefore every covering in $A$ by open subsets of $X$ contains a finite subcover. This is the definition of compact subsets and is ok.


*Every $B\subseteq A$ that is open in $(X,T)$, it is open in $A$ as well (correct, since every element in $T_1$ is in the form $A\cap O$, with $O\in T$, so $B=A\cap B\in T_1$ if $B\in T$).


*"Then $A$ has an open covering in $(A,T_1)$ where the union of the open sets in the covering equals $A$. So $A$ is a compact subset of $(A,T_1)$". The first part is a trivial statement, because it says "$A$ has a cover in $(A,T_1)$ that covers $A$". To reach the conclusion $A$ is compact in $(A,T_1)$ you need to show that any cover for $A$ in $(A,T_1)$ has a finite subcover.
In the second point

*

*The whole initial part "it contains a compact subset $B$ equal to $A$" is superfluous, just say $A$ (without naming $B$).


*"Since the sets in the open coverings of $B$ in $(A,T_1)$ are subsets of sets that form a collection that covers $A$ in $(X,T)$" is correct, but then you say "the intersection of $A$ and another set in $X$ is a subset of $X$", which is trivially true and has nothing to do with the definition of the induced topology $T_1$. Instead, I think you meant "the intersection of $A$ with an open subset of $X$, is an open subset of $A$ in $T_1$", which is the definition of the induced topology $T_1$.


*Finally, you need to show how the point I mentioned above implies that $A$ is compact in $X$, that is, there should be another step between "every cover of $A$ in $(A,T_1)$ corresponds to a covering of $A$ in $(X,T)$" and "$A$ is compact in $(X,T)$". But it is in the right direction, you have to form a correspondence between the covers of $A$ in $(X,T)$ and in $(A,T_1)$
Another way to phrase the proof is the following: If $\mathcal C$ is a open cover of $A$ in $(A,T_1)$, then $$\mathcal C'=\{O\in T|A\cap O\in\mathcal C\}$$ (that is, the opens in $X$ that intersected with $A$ are in $\mathcal C$) is a open cover of $A$ in $(X,T)$. If $A$ is compact in $(X,T)$, then $\mathcal C'$ contains a finite subcover $\{O_1,\cdots,O_n\}$ of $A$, so that $$\{A\cap O_1,\cdots,A\cap O_n\}$$ is a finite subcover of $A$ in $(A,T_1)$ (by definition of $\mathcal C'$, if $O\in\mathcal C'$, then $A\cap O\in\mathcal C$, so $A\cap O_1,\cdots,A\cap O_n$ are elements of $\mathcal C$), so $(A,T_1)$ is compact (we showed a generic cover $\mathcal C$ of $A$ in $(A,T_1)$ has a finite subcover).
On the other hand, if $\mathcal D$ is a cover of $A$ in $(X,T)$, then $$\mathcal D'=\{A\cap O|O\in\mathcal D\}$$ is a open cover of $A$ in $(A,T_1)$. In particular, if $(A,T_1)$ is compact, then $\mathcal D'$ contains a finite subcover $\{A\cap O_1,\cdots,A\cap O_n\}$ and $\{O_1,\cdots,O_n\}\subseteq\mathcal D$ is a finite subcover of $A$ in $(X,T)$, so $A$ is a compact subset of $(X,T)$ (we showed that a generic open cover $\mathcal D$ of $A$ in $(X,T)$ contains a finite subcover).
