Let $(X_i, \mathcal{T}_i)_{i \in I}$ be a family non empty topological spaces with basises $(\mathcal{A_i})_{i \in I}$
Prove that $\mathcal{S}:= \{\operatorname{pr}_k^{-1}(A_k) \mid k \in I, A_k \in \mathcal{A_k}\}$ is a subbasis for the product topology $\mathcal{T}$.
My attempt:
Denote $\mathcal{K}$ with the canonical subbasis of the product topology. Then $\mathcal{S} \subseteq \mathcal{K}$, and hence $\langle \mathcal{S}\rangle \subseteq \langle \mathcal{K}\rangle = \mathcal{T}$, where this notation means "the smallest topology generated by the given set".
Now, define for $k \in I$
$$\operatorname{pr_k}: \left(\prod_{i \in I}X_i, \langle \mathcal{S}\rangle\right) \to (X_k, \mathcal{T}_k): x = (x_i)_i \mapsto x_k$$
Then, if $B \in \mathcal{T}_k$, then $B = \bigcup_{l \in J}A_l$ with $A_l \in \mathcal{A}_l$ for every $l \in J$.
Hence $\operatorname{pr}_k^{-1}(B) = \bigcup_{l \in J}\operatorname{pr}_k^{-1}(A_l) \in \langle \mathcal{S} \rangle$. Hence, $\langle \mathcal{S} \rangle$ makes all the projection continuous and it follows that $\langle\mathcal{S}\rangle \supseteq \mathcal{T}$. Hence, $\mathcal{S}$ is a subbasis for the product topology.
Is this correct?
Yes, this is correct. The essential observation is that taking this subbase makes all projections continuous (this is a general fact: $f: X \to Y$ is continuous, iff for some base of $Y$, inverse images of that base under $f$ are open in $X$), and thus by minimality must equal the product topology.