Is this statement really that technical to prove? Let $\{X_i\}_{i\in I}$ a family of topological spaces and $X=\prod_{i\in I}X_i$ denote the product topology, i.e. a subbasis $\mathcal{C}$ of it is given by
$$\mathcal{C}=\left\{\left.{\textstyle\prod_{i\in I}Y_i}\;\right|\;\exists j\in I:Y_j\text{ is open in }X_j\text{ and }\forall i\in I:i\neq j\Rightarrow Y_i=X_i\right\}$$
Now the basis $\mathcal{B}$ induced by $\mathcal{C}$ is given by
$$\mathcal{B} = \left\{\left.{\textstyle \bigcap_{A\in\mathcal{A}}A}\;\right|\; \mathcal{A}\subseteq\mathcal{C}\text{ is finite}\right\}$$
Given the description of $\mathcal C$ it "immediately" follows that
$$\mathcal{B}=\left\{\left.{\textstyle\prod_{i\in I}Y_i}\;\right|\;\exists J\subseteq I\text{ finite, }\forall i\in I: Y_i\text{ is open in }X_i\text{ and }\left(i\notin J\Rightarrow Y_i=X_i\right)\right\}$$
But working the details of the proof out, I noted that it is rather complicated if done rigorous. Is there an easy way to proof the equality above (rigorously)?
For example, let $\mathcal A\subseteq \mathcal C$ be finite. For each $A\in\mathcal A$ there is an $j\in I$ such that for all $i\in I$ with $i\neq j$ holds $\operatorname{pr}_i A = X_i$, denote this* function by $g:\mathcal A\to I$ (*$g$ isn't unique when $X\in\mathcal A$). $g$ basically denotes the "dimension" where we don't have the full set (unless $A=X$ or $A=\emptyset$). $g$ isn't necessarily injective, set
$$\mathcal A'=\left\{\left.\bigcap_{A\in g^{-1}\{i\}}A\;\right|\;i\in I:g^{-1}\{i\}\neq\emptyset\right\}$$
$\mathcal A'$ is finite and we have $\bigcap_{A\in\mathcal A}A=\bigcap_{A\in\mathcal A'}A$. On the other side, if we define $g':\mathcal A'\to I$ analogously to $g$ above, now $g'$ is injective (given $X,\emptyset\notin\mathcal A$). Furthermore, for each $i\in I$ with $g^{-1}\{i\}\neq\emptyset$ exists a $Z_i$ such that
$$\bigcap_{A\in g^{-1}\{i\}}A=\prod_{j\in I}Z_j^i$$
with $Z_i^i=Z_i$ (open in $X_i$) and for $j\neq i:Z_j^i=X_j$. We finally have
$$\bigcap_{A\in \mathcal A} A = \bigcap_{i\in\operatorname{rng}g}\bigcap_{A\in g^{-1}\{i\}}A=\prod_{i\in I}Y_i$$
with $Y_i=Z_i$ for $i\in\operatorname{rng} g$ and $Y_i=X_i$ else. Setting $J=\operatorname{rng}g$ and noting that $J$ is finite because $\mathcal A$ is finite, we finish the first inclusion. 
Even with skipping some smaller steps and special cases, this is still quite long. While I'm aware intuitive right things can have long proofs (see measure theory), I'm quite surprised if this would be the case here, because the equality above is usually taken as clear in the books I read without mentioning that the proof is technical (as I've seen it e.g. in measure theory).
Edit: In conclusion, I used a different way avoiding induction while the solution from the accepted answer used it (as some books I consulted did, too). The difficulty is masked with the induction, because when you get rigorous, it one of these easy-to-see but hard-to-formalize proofs. But during my research I couldn't find a third way, so I'll just assume there is no easier way.
 A: It's mostly a matter of notation. Letting $\pi_i$ be the projection from $\prod_{j \in I} X_j$ to $X_i$, for all $i \in i$, I would denote the subbase as 
$$\mathcal{C} = \{ \pi_i^{-1}[O_i]: i \in I, O_i \text{ open in } X_i\}$$
and the induced base, which indeed consists of finite intersections of members of $\mathcal{C}$ then is all sets of the form $\cap_{j \in F} \pi_j^{-1}[O_j]$, where $F$ is a finite set of indices, and all $O_j$ are open in $X_j$. If we were to choose two of such sets with the same index $j$, we could put them together in this form too, as $\pi_j^{-1}[O_j] \cap \pi_j^{-1}[O'_j] = \pi_j^{-1}[O_j \cap O'_j]$ etc. So we can assume WLOG that we have different indices $j$.
Then note that  $\cap_{j \in F} \pi_j^{-1}[O_j] = \prod_{i \in I} Y_i$ where $Y_i = X_i$ if $j \notin F$ and $Y_i = O_i$ for $i \in F$. 
Proof: suppose $x \in \cap_{j \in F} \pi_j^{-1}[O_j]$ then for all $i$ we of course have $x_i \in X_i$ but if $i \in J$ we have that $x \in \pi_j^{-1}[O_j]$ so by definition $x_j = \pi_j(x) \in O_j$. So $x$ is in the product set on the right by definition. If $x$ is in that product set, and $j \in F$ then $x_j \in O_j$ and this says $x \in \pi_j^{-1}[O_j]$, and as this holds for all $j \in F$, $x \in \cap_{j \in F} \pi_j^{-1}[O_j]$, so that shows the other inclusion.
So the finite intersections of the subbasic elements form the canonical product base sets as you also described them. I don't see any technical difficulties here...
