This is an example of the book Topology of Dugundji.
Ex. 4 $\,$ It is useful to observe that if $ V $ is any set open in $ \prod_{\alpha} Y_{\alpha} $, then $ p_{\alpha} (V) = Y_{\alpha} $ for all but at most finitely many $ \alpha $: any such $ V $ contains a basic $ U = \langle U_{\alpha_{1}} \cdot \cdot \cdot, U_{\alpha_{n}} \rangle $ and $ p_{\alpha} (U) \subset p_{\alpha}(V)$ for all $ \alpha $.
I understand: "any such $ V $ contains a basic $ U = \langle U_{\alpha_{1}} \cdot \cdot \cdot, U_{\alpha_{n}} \rangle $ and $ p_{\alpha} (U) \subset p_{\alpha}(V)$ for all $ \alpha $". But how do I conclude that $ p_{\alpha} (V) = Y_{\alpha} $ for all but at most finitely many $ \alpha $?
Recall that the basic open sets in the product topology are the product of finitely many open sets with the whole spaces, i.e., if $U$ is an open set of $\displaystyle\prod _{j\in J} Y_j$ then $U=U_1\times\dots U_n\times\displaystyle\prod _{j\notin\{1,...,n\}}Y_j$ where $U_i\subseteq Y_i$ is an open set of $Y_i$. Other way to see $U$ is formed by the finite intersection of the inverse image of open sets under the projection (here the projection is denoted by $\Pi_j$).
Your question follows from here. If $V$ is open, then take $U$ a basic open set such that $U\subseteq V$. Then, for all $j\in J$ we have that $\Pi_j[U]\subseteq \Pi_j[V]$ and, by the previous argument, $U=U_1\times\dots U_n\times\displaystyle\prod _{j\notin\{1,...,n\}}Y_j$. Then, for all $j\notin\{1,...,n\}$ we can conclude that $Y_j=\Pi_j[U]\subseteq\Pi_j[V]\subseteq Y_j$.