I am trying to show that the product space of some connected spaces is also connected, and I am not sure whether I am right or not. Here is my proof:
Proof : Suppose that $\left\{X_\lambda\right\}_{\lambda\in\Lambda}$ is a family of connected spaces and $X_0=\Pi X_\lambda$ is the product space. Denote the $\lambda$'th projection by $P_\lambda$, and by the axiom of choice, $P_\lambda$ is surjective. Suppose that $X_0$ is not connected, then $X_0=U\cup V$, where $U,V$ are disjoint non-void clopen sets, then certainly $P_\lambda\left(U\right)$ and $P_\lambda\left(V\right)$ are both non-void for all $\lambda\in\Lambda$. Then if for all $\lambda\in\Lambda$, $P_\lambda\left(U\right)\cap P_\lambda\left(V\right)\neq \varnothing$, there will be some $x\in X$ such that $P_\lambda\left(x\right)\in P_\lambda\left(U\right)\cap P_\lambda\left(V\right)$ for every $\lambda\in\Lambda$, which implies that $x\in U\cap V$ and contradicts the hypothesis that $U$ and $V$ are disjoint. Thus there shall be some $\lambda_0\in\Lambda$ such that $P_{\lambda_0}\left(U\right)\cap P_{\lambda_0}\left(U\right)= \varnothing$, and since $P_{\lambda_0}$ is surjective, $P_{\lambda_0}\left(U\right)\cup P_{\lambda_0}\left(V\right)= X_{\lambda_0}$. Since $P_{\lambda_0}$ is open, both $P_{\lambda_0}\left(U\right)$ and $ P_{\lambda_0}\left(V\right)$are open and thus are both clopen, which contradicts the connectedness of $X_{\lambda_0}$.
Am I right? Could you please help me?
The proof is not right.
From the assumption
you incorrectly conclude that
To see why this is wrong, consider the subsets
$$\begin{align*} U & = \{ (x, y) \in \mathbb{R}^2 : y < x \} \\ V & = \{ (x, y) \in \mathbb{R}^2 : y \geqslant x \} \\ \end{align*}$$
of the product $\mathbb{R} \times \mathbb{R}$. Clearly $(x, y) = (0, 0) \in \mathbb{R}^2$ is a point satisfying $x \in \pi_x(U) \cap \pi_x(V)$ and $y \in \pi_y(U) \cap \pi_y(V)$, but $(x, y) \notin U$.