Suppose that $F$ is a family of maps from a set $X$ to spaces $Y_f$ indexed by $F$. Let $Y=\prod_{f\in F}Y_f$. Let $P:X\to Y:x\mapsto\langle f(x):f\in F\rangle$ be the evaluation map. One way to show that the weakest topology making each $f\in F$ continuous is the same as the weakest topology making $P$ continuous is to show that a topology $\tau$ on $X$ makes $P$ continuous iff it makes each $f\in F$ continuous.
Suppose that $\tau$ is a topology on $X$ making each of the maps $f$ continuous, so that for each $f\in F$ and each open set $U\subseteq Y_f$, $f^{-1}[U]\in\tau$. Then $U$ is the union of basic open sets of the form
$B=\{y\in Y:\forall f\in F_0(y_f\in U_f)\}\;,\tag{1}$
where $F_0$ is a finite subset of $F$, and $U_f$ is an open set in $Y_f$ for each $f\in F_0$. Thus, to show that $P^{-1}[U]\in\tau$, it suffices to show that $P^{-1}[B]\in\tau$ for each $B$ of the form $(1)$. But
$\begin{align*}x\in P^{-1}[B]&\text{ iff }P(x)\in B\\ &\text{ iff }P(x)_f\in U_f\text{ for each }f\in F_0\\ &\text{ iff }f(x)\in U_f\text{ for each }f\in F_0\\ &\text{ iff }x\in\bigcap_{f\in F_0}f^{-1}[U_f]\;, \end{align*}$
so $P^{-1}[B]=\bigcap_{f\in F_0}f^{-1}[U_f]\;.$
By hypothesis each of the sets $f^{-1}[U_f]\in\tau$, so $P^{-1}[B]\in\tau$ as well, and so is any union of such sets.
Conversely, suppose that $\tau$ makes $P$ continuous. Then $P^{-1}[B]\in\tau$ for every basic open set in $Y$ of the form $(1)$. In particular, for each $f\in F$ and any open set $U_f\subseteq Y_f$, $B=\{y\in Y:y_f\in U_i\}$ is open in $Y$, and therefore $P^{-1}[B]\in\tau$. But
$\begin{align*} P^{-1}[B]&=\{x\in X:P(x)\in B\}\\ &=\{x\in X:P(x)_f\in U_f\}\\ &=\{x\in X:f(x)\in U_f\}\\ &=f^{-1}[U_f]\;, \end{align*}$
so $f^{-1}[U_f]=P^{-1}[B]\in\tau$.
This shows that the topologies making $P$ continuous are exactly the same as the topologies making every map in $F$ continuous, and it follows immediately that the weakest topology making $P$ continuous is the weakest topology making every map in $F$ continuous.