$\Pi_{i \in I}X_i$ Hausdorff $\Rightarrow$ each $X_i$ Hausdorff. Consider a collection of non-empty topological spaces $(X_i)_{i \in I}$ and the product topology $\Pi_{i \in I}X_i$.
I'm trying to prove that if $\Pi_{i \in I}X_i$ is Hausdorff then each $X_i$ is Hausdorff.
I've started similarly to this post: Theorem 19.4 in Munkres' TOPOLOGY, 2nd ed: Does the converse also hold?. However, I have a hard time writing the last steps in my proof.
Here's a sketch of my idea. So following the construction in the link, we have two points in the product topology that are equal in each component except, say the $i$th one. Because $\Pi_{i \in I}X_i$ is Hausdorff, we can find disjoint opens, resp. $U,V$, around these points. I then thought to use the base of the product topology that tells us that we can find $\Pi_{i \in I}U_i \subseteq U$ and $\Pi_{i \in I}V_i \subseteq V$ where $U_i$ and $V_i$ are opens in $X_i$ and $U_i = X_i$ and $V_j = X_j$ for all but finite many $i$ and $j$. Now, somehow we have to have that $U_i \cap V_i = \emptyset$ for the $i$ where the points in the product topology differ.
I am stuck on how to figure this last step out. Also is the step of using the base of the product topology really necessary? I feel like I don't really use it here, but on the other hand don't I need it to get to the $U_i$ and $V_i$?
 A: You are on the right track. Suppose that  $x_i,y_i$ in $X_i$ are distinct, and
fix some $x_j \in X_j$ for all $j \in I\setminus {j}$. then $x=(x_j)_{j \in I} \in X=\prod_j X_j$ and $y$ obtained from $x$ by replacing $x_i$ by $y_i$ and leaving the other coordinates unchanged, can be separated in $X$ by open sets $U,V$. Choose basic open neighborhods $\prod_j U_j \subset U$ of $x$ and $\prod_j V_j \subset U$ of $y$. Note that $x_j \in U_j \cap V_j$ for all $j \ne i$. Now we claim $U_i \cap V_i=\emptyset$. Indeed, if there was some $z_i \in U_i \cap V_i$, then the point
$z$ obtained from $x$ by replacing $x_i$ by $z_i$ and leaving the other coordinates unchanged, would be in $U \cap V$, a contradiction.
A: As noted in comments your approach has issues. Lets try it differently.
One of the ideas is to show that each $X_j$ is homeomorphic to some subspace of $\prod_{i\in I}X_i$ and thus it has to be Hausdorff.
First of all recall how the product space is defined:
$$\prod_{i\in I}X_i=\{f:I\to \bigcup_{i\in I} X_i\ |\ f\text{ is a choice function}\}$$
with appropriate (product) topology.
Now fix a point $e_j\in X_j$ for each $j$ and consider the following function:
$$\lambda_j:X_j\to\prod_{i\in I}X_i$$
$$\lambda_j(x)(i)=\begin{cases}
x&\text{if }i=j\\
e_i&\text{otherwise}
\end{cases}$$
This looks complicated, but for example if we have $X_1\times X_2\times X_3$ then $\lambda_1(x)=(x,e_2,e_3)$, $\lambda_2(x)=(e_1,x,e_3)$ and $\lambda_3(x)=(e_1,e_2,x)$. In other words $\lambda_j$ puts $x$ on $j$-th coordinate, and every other coordinate is fixed.
I leave as an exercise that each $\lambda_j$ is continuous, which easily follows from the universal property of product topology, because the composition $\pi_k\circ\lambda_j$ with each projection is either the identity or a constant map. Moreover each $\lambda_j$ is a homeomorphism onto its image, because the inverse is projection $\pi_j$ restricted to the image.
