Confused: the set of all interior points for $E$ is always open This is the proof my professor used to prove that $E^0$ = {set of all interior points of $E$} is always open.
Proof: Suppose $p \in E^0$. Since $p$ is an interior point of $E$, then, for some $r > 0$, there is a neighborhood $N_r (p) \subseteq E$. We know that neighborhoods are open sets; this means that $\forall q \in N_r (p)$, there is some radius $\epsilon > 0$ s.t. $N_\epsilon (q) \subseteq N_r (p) \subseteq E$. By transitivity, $N_\epsilon (q) \subseteq E$. This means that every point in $N_r (p)$ is an interior point of $E$, which means that $N_r (p) \in E^0$, and $E^0$ is always open.
This is the part I'm confused about:
Why does $N_\epsilon (q) \subseteq E$ mean that every point in $N_r (p)$ is an interior point of $E$? I'm not seeing the connection. If anyone can "dumb down" the explanation for me so I can understand it better, that'd be tremendously helpful. Thank you!

For all $q$ which is in $N_r(p)$, we know that there is $\varepsilon$ such that $N_\varepsilon(q) \subseteq E$, i.e. $q$ is an interior point of $E$.
For any $q$ in $N_r(p)$, $q$ is an interior point of $E$.
Every point in $N_r(p)$ is an interior point of $E$.
