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I've seen references to this statement but not a proof. I've worked up a proof myself and would appreciate confirmation or correction.....

Let $f: (X, d_x) \to (Y, d_y)$ be uniformly continuous on $X$ and $S$ be a totally bounded subset of $X$.

Pick any $\epsilon > 0$, then by uniform continuity there is (a common) $\delta > 0$ with $d_y(f(x_1), f(x_2)) < \epsilon$ whenever $d_x(x_1, x_2) < \delta$. So for any $x \in X$ it follows that $f(B_\delta(x)) ⊂ B_\epsilon(f(x))$

With $S$ totally bounded there is a finite cover of open balls with the determined $\delta$ as the radius, and centred on points $\{x_i\}_{ i = 1, n}$ with each $x_i \in S$.

So, $S \subset \bigcup_{i = 1, n} B_\delta(x_i)$, and therefore $f(S) \subset f(\bigcup_{i = 1, n} B_\delta(x_i)) = \bigcup_{i = 1, n} f(B_\delta(x_i))$.

But each $f(B_\delta(x_i)) \subset B_\epsilon(f(x_i))$ and therefore $f(S) \subset \bigcup_{i = 1, n} B_\epsilon(f(x_i))$.

I.e. $f(S)$ has a finite cover $\{B_\epsilon(f(x_i)) \}_{i = 1, n} $ of open balls with radius the given $\epsilon$ and with corresponding centres $f(x_i) \in f(S)$, which defines $f(S)$ as totally bounded.

As a final point, I think that by considering the subspace topology of $S$ It suffices for $f$ to be uniformly continuous on $S$ and not necessarily on all of $X$. True or false ?

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    In standard usage $\cup$ is used in expressions like $A\cup B$ and $A_1\cup \cdots \cup A_n$ and $\bigcup$ is used in expressions like $\bigcup_{i=1}^n B(x_i)$ or $\displaystyle \bigcup_{i=1}^n B(x_i).$ I changed it. – 2017-01-21
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    @MichaelHardy. Thanks Michael, any feedback on the question itself ? – 2017-01-21
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    The proof looks correct (although I haven't gone through it with the most fine-toothed comb that I have....). As for subspace topologies, with metric spaces the topology on $S$ that comes from restricting the metric to pairs of points in $S$ is the same as the subspace topology. So it's just as if $S$ had been the whole metric space you were considering. – 2017-01-21

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