From Wikipedia, I noticed that the wedge sum is the quotient topology of the disjoint union of two spaces, namely
$$X\vee Y = (X\sqcup Y)/\sim$$
by the identification $$x_0 \sim y_0$$
My question is: Is the wedge sum of any two hausdorff spaces hausdorff? How can I prove/disprove such a claim? I'm not super familiar with what happens to neighborhoods in the disjoint union.
Sorry if my english isnt great
You could do it directly from the definition, but that way is a bit tedious and requires considering many different cases (when neighbourhoods contain or not the special points). There is a faster (I think) way, if you already know few tools.
So first of all, if $X$ and $Y$ are Hausdorff then so is their disjoint union $X\sqcup Y$ (which I leave as a simple exercise). We will now use the following:
Lemma. If $S$ is Hausdorff then $S/\sim$ is Hausdorff if and only if $\sim$ is a closed subset of $S\times S$.
applied to $S:=X\sqcup Y$. So what exactly our $\sim$ is? You can check it manually that in case of the wedge sum we have
$$\sim=\Delta\cup\{(x_0,y_0),(y_0,x_0)\}$$
where $\Delta$ is the diagonal $\{(\alpha,\alpha)\ |\ \alpha\in X\sqcup Y\}$. The diagonal $\Delta$ is closed by the same lemma applied to $S/\Delta$ (which is Hausdorff because it is trivially homeomorphic to $S$). Moreover $(X\sqcup Y)\times (X\sqcup Y)$ is of course Hausdorff, in particular points are closed. And so our $\sim$ is the union of the diagonal and two points. Thus it is closed and therefore $(X\sqcup Y)/\sim$ is Hausdorff.
