Another from Rudin. Just wanting feedback on conciseness, correctness, and style.
Prove $E^\mathrm{o}$ is always open.
Let $p \in E^\mathrm{o}$. Find $r >0$ such that $N_r(p) \subset E$.
We know neighborhoods are open.
So, $\forall q \in N_r(p), \exists r' > 0$ such that $N_{r'}(q) \subset N_r(p) \subset E$. Thus $q$ is an interior point of $E$ and belongs to $E^\mathrm{o}$.
Thus $N_r(p) \subset E^\mathrm{o}$.