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Yes, a subspace can be compact. If $Y$ is a subset of $X$ equipped with the relative topology $\tau_Y$, then $(Y,\tau_Y)$ is itself a topological space.

Conditions: $Y$ must be regarded as a subspace of $X$.; $\tau_Y$ is the relative topology on $Y$.; The same compactness criterion is applied to covers of $Y$.

Yes, a subspace can be compact. If $Y$ is a subset of $X$ equipped with the relative topology $\tau_Y$, then $(Y,\tau_Y)$ is itself a topological space.

Conditions: $Y$ must be regarded as a subspace of $X$.; $\tau_Y$ is the relative topology on $Y$.; The same compactness criterion is applied to covers of $Y$.

A topological space $(X,\tau)$ is compact if every open cover of $X$ has a finite subcover. This means that for any collection of open sets whose union contains $X$, there exists a finite subcollection of those open sets that still covers $X$.

Conditions: The object is a topological space $(X,\tau)$.; The quantifier is over every open cover of $X$.; The required conclusion is the existence of a finite subcover.

A topological space $(X,\tau)$ is compact if every open cover of $X$ has a finite subcover. This means that for any collection of open sets whose union contains $X$, there exists a finite subcollection of those open sets that still covers $X$.

Conditions: The object is a topological space $(X,\tau)$.; The quantifier is over every open cover of $X$.; The required conclusion is the existence of a finite subcover.