Let $\varphi: X \to Y$ be a continuous map between two Hausdorff topological spaces $X$ and $Y$. Let $\varphi(x) = y$ for $x \in X$, $y \in Y$ and $U$ be an open neighborhood of $y$.
Question: Does there exist an open neighborhood $U'$ of $x$ such that $\varphi(U') \subseteq U$?
Motivation: I need it for the proof that the inverse limit of Hausdorff topological spaces is a closed subspace of the underlying product space. The question above is exactly the one I have in the proof shown in Lemma 1.1.2 in the book Profinite Groups by Ribes and Zalesskii, cf. this link.
Thoughts and Ideas: Considering images from the Euclidean spaces, I think one must use the continuity at least one for an argument. I also thought that one could choose $U' = \varphi^{-1}(U)$ which should be open (since $\varphi$ is continuous) and $x \in U'$ (by definition of the inverse image). I thought that this argument should be correct, but I was wondering why Ribes/Zalesskii made it so complicated then.
Could you please verify my line of reasoning? Thank you!
To answer your question, yes.
That is a direct result of the theorem
f:X -> Y is continuous at a iff
for all open V nhood f(a), exists open U nhood x with f(U) subset V.