Suppose I have two sets $P, Q \subseteq R^d$ such that $P\subseteq Q$. $P$ and $Q$ are both convex and closed. I wish to show that $P=Q$.
A straightfoward way to show this is showing $\forall y \in R^d$, $y \notin P \Rightarrow y \notin Q$. However, I follow a slightly different technique which utilizes the closedness and convexity of both the sets which I believe is known as the continuity method in topology (although I may be wrong.)
I can show in my problem that for any $x \in P$, $\exists \epsilon$ such that $\forall y \in B(x,\epsilon)$ (closed ball of radius $\epsilon$ around a point $x \in R^d$) I have, $y \notin P \Rightarrow y \notin Q$ i.e. all points immediately outside the boundary of $P$ don't belong to $Q$ as well.
I believe this is sufficient to show $P=Q$. Is this fine? If yes, does anyone have a reference I can cite for this (this is probably not very deep but the publication is applied and hence a reference is needed.) Reference at the level of appropriate theorems would be most appreciated.