$1)$ If $(x_n)$ is Cauchy in $(X,d)$, then $x_n \longrightarrow x_0$. Or, equivalently, $\forall \epsilon > 0, \exists N \in \mathbb{N}$ s.t: $d(x_n,x_0) < \epsilon , \forall n \geq N$.
$2)$ Defn: A function $f: X \longrightarrow Y$ is called uniformly continuous if, $\forall \epsilon > 0, \exists \delta > 0$ s.t: $\rho(f(x),f(y)) < \epsilon, \forall x,y \in X$ satisfying $d(x,y) < \delta$.
Proof: It suffices to show that $(f(x_n))$ converges to some $f(x_0)$ in $(Y,\rho)$. Indeed, since $(x_n)$ is Cauchy, it converges in $(X,d)$ and so $\forall \delta > 0, \exists N \in \mathbb{N}$ s.t: $d(x_n,x_0) < \delta$, whenever $n \geq N$. So fix $\delta > 0$. Then, in view of the uniform continuity of $f$, we have that $\forall \epsilon > 0, \exists \delta > 0$ s.t: $\rho(f(x_n),f(x_0)) < \epsilon, \forall x_n,x_0 \in X$ satisfying $d(x_n,x_0) < \delta$.
In other words, $(f(x_n)) \longrightarrow f(x_0)$, and so $(f(x_n))$ is a Cauchy sequence in $(Y,\rho)$ since it converges. QED.
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