3 added 456 characters in body

Alternatively, you may use the function $f:\mathbb{R\rightarrow\mathbb{R}}$ , expressed by $f\left(x\right)=x^{2}$ to show that it is possible for a continuous function to send Cauchy sequences to Cauchy sequences without being uniformly continuous.

As for the continuity question: the function must be continuous. For, embed $Y$ into its completion, say $Y^{\sim}$, let $p\in X$ , and let $\left(x_{n}\right)$ be a sequence in $X$ converging to $p$ . Then the sequence $(x_{1}, p, x_{2}, p,...)$ is Cauchy in $X$ , isn't it ? And, therefore, its image by $f$ should be a Cauchy sequence in $Y$ , hence convergent in $Y^{\sim}$ . Yet, that image contains a constant subsequence... isn't it ?

2 added 257 characters in body

Alternatively, you may use the function $f:\mathbb{R\rightarrow\mathbb{R}}$ , expressed by $f\left(x\right)=x^{2}$ to show that it is possible for a continuous function to send Cauchy sequences to Cauchy sequences without being uniformly continuous.