Proposition 6.4.2.label Let $X, Y$ be uniform spaces and $f \in UC(X; Y)$. If $X$ is totally bounded, then so is $f(X)$.
Proof. Let $U$ be an entourage of $Y$, then there exists $\seqf{x_j}\subset X$ such that $\bigcup_{j = 1}^{n}[(f \times f)^{-1}(U)](x_{j}) = X$. In which case, $f(x) \subset \bigcup_{j = 1}^{n} U(f(x_{j}))$.$\square$
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