Let $(X,d), (Y,\delta)$ be metric spaces. Then we know the following:
Proposition $1$. A function $f:X\rightarrow Y$ is continuous (by definition, reflects open sets to open sets) iff maps convergent sequences to convergent sequences.
Proposition $2$. If a function $f:X\rightarrow Y$ is uniformly continuous (by $\epsilon-\delta$ definition), then it is continuous.
Proposition $3$. If a function $f:X\rightarrow Y$ is uniformly continuous, then it maps Cauchy sequences to Cauchy sequences. The converse is false: given that $\mathbb{R}$ is a complete metric space, there are continuous functions $f:\mathbb{R}\rightarrow \mathbb{R}$ which are not uniformly continuous, such as $e^x$.
Indeed, if the converse of Prop. $3$ were true, then we would get an interesting result, by rephrasing Prop. $2$:
Proposition. $4$. If a function $f:X\rightarrow Y$ maps Cauchy sequences to Cauchy sequences, then it does the same with convergent ones.
Due to this, the converse of Prop. $4$ is false.
QUESTION. Now, if $Y$ is also complete, then Prop. $4$ is obviously true. How about in general? If it is true, then we have a new form of continuity (preserving Cauchy sequences), between continuity and uniform continuity. I suppose it is false, but I can't come up with a counterexample, since I don't know many metric spaces which are not complete. Thank you in advance!
Prop 4 is true. Suppose $f$ maps Cauchy sequences to Cauchy sequences, and let $x_n$ be a sequence that converges to $x$. Consider the sequence $x_1, x, x_2, x, x_3, x, \ldots$ consisting of $x_n$ with $x$ interspersed. This is also Cauchy, so the sequence $f(x_1), f(x), f(x_2), f(x), f(x_3), f(x),\ldots$ must be Cauchy. But that implies $f(x_n)$ converges to $f(x)$.