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**Bruno J.** You're right! Here's my proof.

Suppose $\displaystyle D$ is bounded and $\displaystyle f$ takes Cauchy sequences to Cauchy sequences. I claim that $\displaystyle f$ can be extended to a continuous function on $\displaystyle \overline D$, which is compact; then $\displaystyle f$ is uniformly continous on $\displaystyle \overline D$ and thus certainly on $\displaystyle D$ also.

It suffices to show that if $\displaystyle (x_n), (y_n)$ are two Cauchy sequences with the same limit $\displaystyle d \in \overline D$, then $\displaystyle (f(x_n)),(f(y_n))$ have the same limit which we may then call $\displaystyle f(d)$. Consider the sequence $\displaystyle (x_1, y_1, x_2, y_2,\dots)$, which converges to $\displaystyle d$ also; by assumption, it is mapped by $\displaystyle f$ to a convergent sequence in $\displaystyle \mathbb{R}^q$, which implies the subsequences $\displaystyle (f(x_n)),(f(y_n))$ have the same limit!