Let $E_1, E_2$ metric spaces and $p_1:E_1\times E_2\rightarrow E_1$, $p_2:E_1\times E_2\rightarrow E_2$ the projection map. How to prove that if $A$ is a open set in $E_1\times E_2$ then $$p_1(A)=\bigcup_{x_x\in E_2}p_1(A\cap(E_1\times \{x_2\}))$$
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This has nothing to do with $A$ being open, or $E_1,E_2$ having metrics. For any map $f:X\to Y$ from set $X$ into set $Y$, and for any collection of subsets $B_\alpha\subseteq X$ it holds that $$f\Big(\bigcup_{\alpha} B_\alpha \Big) = \bigcup_\alpha f(B_\alpha) \tag1$$ The identity in your question is a special case of (1).
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