Let $(Y,\tau)$ and $(X_i,\tau_i)$, $i= 1,2,\ldots,n$ be topological spaces. Further for each $i$, let $f_i$ be a mapping of $(Y,\tau)$ into $(X_i, \tau_i)$. Prove that the mapping
$$f:(Y,\tau)\longrightarrow \prod(X_i,\tau_i),$$
given by $f(y) =\langle f_1(y),f_2(y),\ldots,f_n(y)\rangle$,is continuous if and only if every $f_i$ is continuous. Hint: $f_i=p_i(f)$ where $p_i$ is the projection mapping of the $i$th component.
This seems trivial for both directions so I'm not sure if I'm missing something.
Assume each $f_i$ is continuous. The take an open set $A$ in the product topology where its basis are open sets from each $X_i$. Since $f$ is continuous, the preimage of $A$ in the product topology is an $n$-tuple of open sets in $Y$ which means each component $f_i$ is continuous.
For the other direction, we assume $f$ is open and show each $f_i$ is open. Using the hint, each $f_i=p_i(f)$ which is a composition of two continuous functions since the projection mapping is continuous. Then each $f_i$ has to be continuous.
Best Answer
Your proof is almost correct. If $f = (f_1, \ldots, f_n)$ and all $f_i$ are continuous, then for a basic open subset $O=\prod_{i=1}^n O_i$, where $O_i \subseteq X_i$ is open for all $i$, we have that $$f^{-1}[O] = \{y \in Y\mid f(y) \in \prod_{i=1}^n O_i\}= \{y \in Y\mid \forall i \in \{1, \ldots n\}: f_i(y) \in O_i \} = \bigcap_{i=1}^n f_i^{-1}[O_i]$$
which is a finite intersection (not product or $n$ tuple!) of open subsets by the continuity of the component mappings $f_i$, and hence open. As inverse images under $f$ of basic open subsets are open, the same holds for all open subsets of the product topology and so $f$ is continuous.
The reverse indeed simply follows from $f_i = \pi_i \circ f$ for all $i$ and the fact that compositions of continuous maps are continuous.
It's easy to see how this generalises to infinite products (if you've treated these as well), but not to the box topology on such a product.