I have the next question:
Consider $X$ a topological space that is non-first countable. We can assume that $X$ is Tychonoff.
Is there a way to embed $X$ in a topological space $Y$ such that $X$ is dense in $Y$ but, at least, there exist $x\in Y\setminus X$ such that $x$ has a countable local basis? I tried to emulate the construction of the one-point compactification of a topological space but it fails, or maybe I can't prove it, in the first countability of the point at infinity.
The reason of the question comes from the analogous of compactifications: we start with a non compact space and then, we embed it, as a dense subespace, of a compact space. Any hint? Thanks in advance.
You didn't assume anything about compactness of $X$, so the idea to use compactification cannot be correct, since $X$ might be its own compactification.
Since you didn't assume any property of $Y$ itself, then there is a quite trivial solution. Consider $Y=X\sqcup \{*\}$ with topology given by: $U\subseteq Y$ is open if and only if $U=Y$ or $U\subseteq X$ is open in $X$.
Clearly $X$ embeds into $Y$. Moreover the only open subset containing $*$ is $Y$ itself, and this implies two things: that the local basis is trivially countable, and that $*$ belongs to the closure of $X$, and so $X$ is dense in $Y$.
Note that I didn't use any assumption on $X$, meaning it literally holds for any topological space $X$.