In Willard's General Topology, in the proof of 8.12, page 56 (Dover edition), I find this:
The aim is to prove that the map $e$ is an open map. $e$ is continuous and one-one as an hypothesis, and the author uses explicitly the one-one property (but I don't understand how).
e is continuous. [...] e is one-one. Finally, we will show e is an open map ; i.e., if U is open in X, then e(U) is open in e(X). Since e is one-one, it suffices to show e(U) is open whenever U is a subbasic open set.
How is it important that $e$ is one-one for that? To me if would seem that for any function $e$ we could check the property on a subbase of $U$ in order to prove that $e$ is open. As all the opens are finite intersection of unions of subbase sets in X, their images by $e$ are similar combinations in e(X), so that the open images of these combinations would also be opens in e(X).
What did I miss? Or is it an overlook from the author?
Prove if $\mathcal{C}$ is a collection of sets and $f$ an injective function that
$$f[\bigcap \mathcal{C}] = \bigcap \{ f[C] : C \in \mathcal{C} \}$$
Let $\mathcal{C}$ be a finite collection of subbase sets.
Use the above to show that $f[\bigcap \mathcal{C}]$ is open. And such sets form a base for $X$.