How to prove the product of totally bounded uniform spaces is totally bounded? 
One should note that the family here may not countable. If it is countable, the it is a consequence of the following results;
Lemma 1. the product of countable totally bounded metric spaces is totally bounded.
Lemma 2. Every uniform space is uniformly isomorphic to a product of metrizable uniform spaces.
Lemma 3. The product of uniform spaces is uniform induced.
How about the case when uncountable?
 A: Every factor space embeds into the product as a subspace, so the previous theorem in Engelking's General Topology (which you're quoting here) implies the left to right implication right away.
To check the right to left one, let all $(X_s, \mathcal{U}_s)$ be totally bounded. It suffices to check the definition of total boundedness for a basic entourage for $X=\prod_{s \in S} X_s$ which is of the form $U=\{(x_s,y_s)_{s \in S}: |x_{s_i} - y_{s_i} | < V_{s_i}, i=1,\ldots,N\}$, where $s_1,\ldots, s_N$ are finitely many indices from $S$ and $V_{s_i} \in \mathcal{U}_{s_i}$. For each $i\in \{1,\ldots,N\}$ we can find a finite set $F_i \subseteq X_{s_i}$ that forms is $V_{s_i}$-dense. Fix any point $p \in X$ for "defaultness" and define $$F = \{(x_s) \in S: \forall i \in \{1,\ldots,N\}: x_{s_i} \in F_i \text{ and } \forall s \in S\setminus\{s_1,\ldots,s_N\}: x_s  =p_s\}$$
and note that $|F| = \prod_{i=1}^N |F_i|$ and hence is finite and is the required finite $U$-dense set. 
A: The proof depends on which theorems you allow to use. You certainly can give a direct proof, but in my opinion the most elegant way is this:
Theorems used:
(1) Each uniform space $X$ has a completion $X'$ (i.e. a complete uniform space containing $X$ as a dense subset).
(2) If $X$ is totally bounded, then $X'$ is totally bounded.
(3) A uniform space is compact if and only if it is complete and totally bounded.
Proof:
If $P = \Pi_{s \in S} X_s$ is totally bounded, then each $X_s$ is totally bounded because it embeds as a uniform subspace into $P$.
If all $X_s$ are totally bounded, then the completions $X'_s$ are compact. Therefore $P' = \Pi_{s \in S} X'_s$ is compact. Since $P$ embeds as a uniform subspace of $P'$, it is totally bounded.
