Let $\{(X_i, p_i)\}_{i \in I}$ be a collection of based Hausdorff spaces. Show that $\bigvee_{i \in I}X_i$ is Hausdorff
I tried to prove this but ended up getting stuck. Basically the idea I had was the following. By definition we have that $$\bigvee_{i \in I}X_i = \bigsqcup_{i \in I}X_i / \{p_i \ | \ i\in I\}$$ and we know that $\bigsqcup_{i \in I}X_i$ is Hausdorff. So now pick two points $x, y \in \bigvee_{i \in I}X_i$. Then consider the quotient map $$q : \bigsqcup_{i \in I}X_i \to \bigsqcup_{i \in I}X_i / \{p_i \ | \ i\in I\}.$$
I was thinking that since all quotient maps are open maps I could choose elements $\alpha \in q^{-1}[\{x\}]$ and $\beta \in q^{-1}[\{y\}]$ and then select disjoint open neighborhoods $U$ and $V$ of $\alpha$ and $\beta$ respectively and then show that $q[U] \cap q[V]$ is empty (and that way construct the required neighborhoods to show Hausdorfness). But I ran into a roadblock because I have no way to ensure that $q[U] \cap q[V] = \emptyset$ . For example if $U$ contains a base point $p_j$ and $V$ contains a base point $p_k$ then I'd have $\{[p_i]\} \subseteq q[U] \cap q[V] $.
How can I go about proving the above?
You've got the right idea. Just do it by points. Pick distinct points $a, b$ in your wedge. Suppose neither one is the basepoint. If $a \in X_i$ then we can separate $a$ from $p_i$ by assumption. Do the same for $b \in X_j$. Now it's clear those those open sets around $a$ and $b$ respectively remain open in the wedge (because they avoid all basepoints by design) and are still disjoint. That case is now done.
Now suppose that $a$ is the basepoint in the wedge and $b$ is not. If $b \in X_j$ then separate $p_j$ from $b$ in $X_j$. This is given by, say, open sets $U$ and $V$ in $X_j$ with $p_j \in U$ and $b \in V$. You're then tempted to say that the images of $U$ and $V$ in the wedge do the trick, but that's not so because $q(U)$ probably isn't open: its inverse image contains $U$ and all the basepoints $p_i$ ($i \in I$). Those other basepoints may not be open in their respective $X_i$.
The fix is simple. $q(V)$ is still open in the wedge (and contains the "original" $b$) as it doesn't hit the basepoint, so we're fine there. For the open set about $a$ in the wedge, take $$ \left( \bigcup_{i \neq j}q(X_i) \right) \cup q(U). $$ This clearly contains $a$ and avoids $q(V)$, but is it open? Yes, because its preimage under the quotient map is $\left(\bigcup_{i \neq j} X_i\right) \cup U$ which is open in the disjoint union (before identifying basepoints). The point is that the preimage of $q(U)$ isn't $U$ but is instead $U \cup \{p_i \mid i \in I\}$, but all those basepoints get gobbled up in the open $X_i$, so the total preimage is open as claimed.