I am trying to learn how to prove that the preimage of an open set is open in general topology. Here is an example that I am not really satisfied with
Proposition 3.9 (Book: Essential Topology, Crossley): If $S$ has the discrete topology and $T$ is any topological space, then any function $f:S\to T$ is continuous.
Proof: If $S$ has the discrete topology, then every subset of $S$ is open, so in particular, every preimage of an open set must be open. Thus $f$ is continuous.
Here is my proof: Let $(X,\mathfrak{T}_{X})$ and $(Y,\mathfrak{T}_{Y})$ be topological spaces, where $\mathfrak{T}_{X}$ is the discrete topology, while $\mathfrak{T}_{Y}$ is any topology on $Y$. We wish to show that $U\in \mathfrak{T}_{Y}\implies f^{-1}(U)\in \mathfrak{T}_{X}$.
Assume that $U$ is open in $Y$. Pick $a\in f^{-1}(U)$. Then $f(a)\in U$. Since $U$ is open, there exists neighbourhood $V_{a}$ of $a$ such that $f(V_{a})\subseteq U$. Then $V_{a}\subseteq f^{-1}(U)$. My thought is now to make $f^{-1}(U)$ "completely" open that contains a lot of neighbourhoods. Letting $V=\bigcup_{x\in f^{-1}(U)}V_{x}$, so $V_{a}\subseteq V=f^{-1}(U)$. Hence $f^{-1}(U)$ is open in $X$. $\square$
How is the proof so far? Any improvings or ideas are much appreciated.
This is not necessarily the case. Since $U$ is open, you know a neighborhood of $f(a)$ is contained in $U$. That is not the same thing as saying that the image of a neighborhood of $a$ is contained in $U$.
The real problem, though, is that you have not used anywhere in your proof that the topology on $X$ is the discrete topology. If your proof were correct, it would imply every map between topological spaces is continuous, which is not the case. The proof really is as simple as the one in the textbook. You wish to show that given an open subset $U$ of $Y$, $f^{-1}(U)$ is an open subset of $X$. Well every subset of $X$ is open, so $f^{-1}(U)$ is an open subset of $X$.