Let $X=\prod_{\alpha \in A} X_\alpha$ be the product of topological spaces with the product topology. Pick a fixed $b_\alpha \in X_\alpha$, for each $\alpha$, and let $B_\alpha=\{x \in \prod X_\alpha : x_\beta =b_\beta \text{ unless } \beta = \alpha\}$, for each $\alpha$. Then the restriction to $B_\alpha$ of the projection map $\pi_\alpha$ is an homeomoprhism.
I understand why $\pi_\alpha |_{B_\alpha}$ is a continuous bijection, but I don't know why it's an homeomprhism. If it was open, that would imply that it's an homeomorphism, but the fact that $\pi_\alpha$ is open doesn't imply that $\pi_\alpha |_{B_\alpha}$ is open, unless $B_\alpha$ is open in $\prod X_\alpha$, but I don't know if that's true.
Fix $\alpha$, and let $f:=\left. \pi_a\right|_{B_\alpha}$ be the map from $B_\alpha$ to $X_\alpha$.
Bijectivity of $f$ is straightforward: identify each point of $B_\alpha$ with its $\alpha$-component.
Continuity of $f$ is also straightforward: an open set $U \subset X_\alpha$ has preimage $\prod_\beta Y_\beta$ where $Y_\beta := \begin{cases}U & \beta=\alpha \\ \{b_\beta\} &\beta \ne \alpha \end{cases}$, which is open in $B_\alpha$.
Openness of $f$ is similar: an open set of $B_\alpha$ takes the form $\prod_\beta Y_\beta$ where again $Y_\beta := \begin{cases}U & \beta=\alpha \\ \{b_\beta\} &\beta \ne \alpha \end{cases}$ with $U$ some open set of $X_\alpha$. $f$ maps this open set to $U$ in $X_\alpha$, which is open.