Definition confirmation: cover of a subset

In lectures we defined:

Let $X$ be a topological space, $Y \subset X$ a subset. A collection $\mathcal{A} \subset \mathcal{P}(X)$ is a cover of $Y$ by sets open in X if each element of $\mathcal{A}$ is an open set in $X$ and $Y \subset \cup_{A \in \mathcal{A}}A$.

We then went on to a proposition:

Let $Y$ be a subset of $X$. Then $Y$ is compact if and only if every cover of $Y$ by subsets open in $X$ has a finite subcover.

I think the "subsets" part in the proposition should actually be "sets", but I want to make sure since we could end up with something different otherwise.

• In my view "subsets open in $X$" somehow implicates that we are dealing here with subsets of $X$. According to that interpretation the statement is true. Something else: I would rather use the statement to define compactness of $Y$. – drhab Feb 25 '16 at 21:39
Saying that $B$ is a subset of $X$ does not mean that $B$ is not equal to $X$. (We call those proper subsets.) Hence, it does not make any difference whether you call them sets or subsets.