So I'm pretty sure this is almost immediate from the definitions, please tell me if I am incorrect..
Consider two cauchy sequences in D, $\{x_n\}$ and $\{y_m\}$. Since $f$ is uniform continuous we know that there exists $\delta > 0$ for all $ \epsilon > 0$ such that for $x_n , y_m$ ($\forall n,m)$
$|x_n – y_m| < \delta \implies |f(x_n) – f(y_m)| < \epsilon$
$f(x_n), f(y_n)$ are clearly sequences, and since we have this inequality for all $n,m$ we conclude that $f(x_n) , f(y_n)$ are cauchy sequences.
Best Answer
No, it's not.
You need to start with one Cauchy sequence, say $(x_n)$. Then you need to show that $\bigl(f(x_n)\bigr)$ is a Cauchy sequence.
To do this, first fix an $\epsilon>0$. You now need to show that there is an $N$ so that $\bigl|f(x_n)-f(x_m)\bigr|<\epsilon$ whenever $n,m\ge N$.
To do this, first use the uniform continuity of $f$ to find a $\delta>0$ so that $|x-y|<\delta$ implies $\bigl|f(x)-f(y)\bigr|<\epsilon$. Now use the Cauchyness of $(x_n)$ to find your desired $N$ (you can make $x_n$ arbitrarily close to $x_m$ for $n,m$ sufficiently large; make them closer than $\delta$).