There are two different definitions for a subbase of a topological space that are not equivalent:
Let $(X,\mathcal{T})$ be a topological space.
1) $\gamma \subset \mathcal{T}$ is a subbase of $\mathcal{T}$ if the collection of all finite intersections of elements of $\gamma$ together with the set $X$, forms a basis for $\mathcal{T}$.
2) $\gamma \subset \mathcal{T}$ is a subbase of $\mathcal{T}$ if additionally $\gamma$ covers $X$, i.e. $\bigcup\limits_{U \in \gamma} U=X$.
theorem of alexander: If every open cover by elements from $\gamma$ has a finite subcover, then the space is compact.
question: Does the theorem holds even for the 1) defintion?
considerations: If $\gamma$ doesn't cover $X$, then for each open cover $\mathcal{U}\subset \mathcal{T}$ of $X$ there holds $X \in \mathcal{U}$ (since each open set $U \in \mathcal{U}$ can be expressed as union of finite many intersections of elements of $\gamma$) and hence $X$ is a finite subcover of itself. So $X$ is compact.
Best Answer
Yes because with the addition of X, $\gamma$ will cover X.
The 2nd definition is much preferred to the unusual, weaker 1st definition.