If for every sequence $(x_n)$, $x_n \to x_0 \implies (f(x_n))$ is a Cauchy sequence, then is $f$ continuous?

The statement is correct. To see this, let $x_0\in D$ and $(x_n)$ a sequence in $D$ with $x_n\to x_0$. We want to show that $f(x_n)\to f(x_0)$. Define the sequence $(y_n)=(x_1,x_0,x_2,x_0,x_3,x_0,\dots)$. Then $y_n\to x_0$, so by assumption $f(y_n)$ is a Cauchy-sequence. It has the constant convergent subsequence $f(x_0)$, hence $f(y_n)$ converges to $f(x_0)$ (A Cauchy-sequence with convergent subsequence is itself convergent with the same limit). As $f(x_n)$ is a subsequence of $f(y_n)$ it follows that $f(x_n)\to f(x_0)$