Is this a homeomorphism? Suppose you have a cartesian product of spaces $\prod_{\alpha\in\mathcal{A}}X_{\alpha}$  in the product topology. 
Choose any $\alpha\in\mathcal{A}$ . Is the following a homeomorphism of a subspace of $\prod_{\alpha\in\mathcal{A}}X_{\alpha}$  with $X_{\alpha}$? 
For each point $p_{\alpha}\in X_{\alpha}$  choose one point $(x)\in\prod_{\alpha\in\mathcal{A}}X_{\alpha}$  with $x_{\alpha}=p_{\alpha}$ . Let's call the set of all these points $P$ .
Consider the projection map $\pi_{\alpha}$  restricted to $P$. It is continuous, open and one-to-one, thus a homeomorphism between $P$ and $X_{\alpha}$.
 A: No, that construction will not in general give you a subspace of $X$ homeomorphic to $X_\alpha$. A simple counterexample is the graph of the function
$$f:\Bbb R\to\Bbb R:x\mapsto\begin{cases}
0,&\text{if }x<0\\
1,&\text{if }x\ge 0\;:
\end{cases}$$
it’s not connected, so it can’t be homeomorphic to $\Bbb R$. (Here $|A|=2$, and both factor spaces are $\Bbb R$.)
What does work is to pick any one point $x\in X$ and let
$$Y=\big\{p\in X:p_\beta=x_\beta\text{ for all }\beta\in A\setminus\{\alpha\}\big\}\;;$$ 
then $Y$ is homeomorphic to $X_\alpha$. In the case of the product $\Bbb R^2$ in the example above, for instance, this construction gives you horizontal lines and vertical lines in the plane, which are all homeomorphic to $\Bbb R$.
A: Here is a counterexample: Let $X_1=X_2=\Bbb R$. Thus $X_1\times X_2=\Bbb R^2$ with the metric topology. For $p\in X_1,\ p\ge0$ choose the pair $(p,1)$ and for $p<0$ choose $(p,0)$. Then the projection map $\pi_1$ restricted to $P$ is not a quotient map (hence not open).
