Does showing one open cover with a finite subcover prove compactness for all covers?

A collection $C$ is called an open cover if it is a cover of $X$ and every member of $C$ is an open set in the topology $\tau$. Showing $\bigcup C = X$ only proves it is a cover; membership in $\tau$ proves it is an open cover.

Conditions: Every member of $C$ belongs to $\tau$.; $\bigcup C = X$.

To check whether a collection $C$ is a cover of $X$, form the union of all members of $C$. If the union equals $X$ (or contains $X$), then every point of $X$ lies in at least one member of $C$, so $C$ is a cover.

Conditions: $C$ is a collection of subsets of $X$.; The test is whether $\bigcup C = X$.

The video concludes $X$ is compact because it demonstrates that an open cover $C$ of the finite space $X$ has a finite subcover $C_1$. The speaker applies the definition of compactness (every open cover has a finite subcover) to this finite setting.

Conditions: $X$ is finite.; Compactness is defined by the requirement that every open cover of $X$ has a finite subcover.

To find a finite subcover, select a finite subcollection $C_1 \subseteq C$ such that the union of the sets in $C_1$ still equals $X$. The chosen sets must come from the original cover $C$.

Conditions: $C_1$ must be a subcollection of $C$.; $C_1$ must be finite.; The union of $C_1$ must equal $X$.

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 $Y$.