Problem Statement:
Let $((E_i,T_i))_{i\in I}$ be a family of non-empty topological spaces. If $U$ is any $\Pi_{i\in I} T_i$-open subset of $\Pi_{i\in I}E_{i}$ prove that $\pi_{i}(U)=E_i$ for all but finitely many indices $i$ in $I$, where $\pi_i$ is the projection mapping from $E$ to $E_i.$
My Attempt:
If $U$ is any open set in the product topology then we have that $$U=\bigcup_{k\in K}\prod_{i\in I}U_{ik}$$ where $U_{ik}\in T_{i}.$ Thus $$\pi_{i}(U)=\pi_{i}\left(\bigcup_{k\in K}\prod_{j\in I}U_{jk}\right)$$ $$=\bigcup_{k\in K}\pi_{i}\left(\prod_{j\in I}U_{jk}\right)=\bigcup_{k\in K}U_{ik}.$$ How do I proceed after this step?
Your approach is correct, but you forgot that for each basic open $B = \Pi_{i \in I} U_i$ you have $U_i = E_i$ for all but finitely many $i$. This means that $\pi_i(B) = E_i$ for all but finitely many $i$. Therefore, if $B \subset M$, then $\pi_i(M) = E_i$ for all but finitely many $i$.