In "Introduction to Topological Manifolds" by John M. Lee, the following in stated in the proof of Theorem 5.10 pp. 102.
Note that $\text{Int } e\cap \text{Int } e'$ is open in $\text{Int } e$. On the other hand, $e'$ is a compact subset of the Hausdorff space $M$, so it is closed in $M$, and therefore $\text{Int } e\cap\text{Int } e'=\text{Int } e\cap e'$ is closed in $\text{Int } e$.
Why must $\text{Int } e\cap \text{Int } e'$ be open in $\text{Int } e$?
In the subspace topology, the open sets are intersections of opens in the ambient space with the subspace. $\text{Int } e'$ is open in $M$ because it's an interior. Hence by definition of subspace topology, $\text{Int } e'\cap \text{Int } e$ is open in $\text{Int } e$.