Let $\mathscr{B}=\left\{\prod_{\alpha\in\lambda}U_\alpha:U_\alpha\in T_\alpha\text{ for each }\alpha\in\lambda\right\}\;;$ clearly is $\mathscr{B}$ is a base for $T$, and for each $B\in\mathscr{B}$ we have $B=\prod_{\alpha\in\lambda}\pi_\alpha[B]\;,$ and $\pi_\alpha[B]\in T_\alpha$ for each $\alpha\in\lambda$.
Fix $\alpha\in\lambda$, and let $U\in T$ be arbitrary. There is a $\mathscr{B}_U\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{B}_U$. Then
$\pi_\alpha[U]=\pi_\alpha\left[\bigcup\mathscr{B}_U\right]=\bigcup\big\{\pi_\alpha[B]:B\in\mathscr{B}_U\big\}\in T_\alpha\;,$
since $\pi_\alpha[B]\in T_\alpha$ for each $B\in\mathscr{B}$.