$f$ is uniformly continuous on $X$ if and only if $d(x_n,z_n)\to0\implies\rho(f(x_n),f(z_n))\to0.$ Let $(X,d)$ and $(Y,\rho)$ be metric spaces and $f:X\to Y$.
Prove that $f$ is uniformly continuous on $X$ if and only if for any sequences $(x_n)$ and $(z_n)$ in $X$,
$$d(x_n,z_n)\to0\implies\rho(f(x_n),f(z_n))\to0.$$
I know that $f$ is uniformly continuous on $X$ if $\forall\epsilon>0, \exists\delta>0$ such that $d(x,x')<\delta\implies\rho(f(x),f(x'))<\epsilon$. It seems that we can apply it directly to prove the question by changing $x$ with $x_n$ and $x'$ with $z_n$ and choose $\delta=\epsilon$ since it converges to 0. I am not sure whether my reasoning is correct? Can anyone please lend me some help?
Also, I am tempted to use Cauchy, but it is not exactly Cauchy.
Any ideas how to write the proof?
Thank you for the helps!

The direction "uniform continuity implies the sequence property" is a bit sloppy. 
Suppose $d(x_n, z_n) \rightarrow 0$. We want to show that $\rho(f(x_n), f(z_n)) \rightarrow 0$. So pick $\varepsilon > 0$. Then pick $\delta > 0$ from the definition of uniform continuity of $f$. Then there is some $N$ such that 
$$\forall n \ge N: \left| d(x_n, z_n) - 0 \right| = d(x_n, z_n) < \delta $$
so for that same $N$
$$\forall n \ge N: \rho(f(x_n), f(z_n)) = \left| \rho(f(x_n), f(z_n)) - 0 \right| < \varepsilon$$
as required.
Now suppose the sequence condition holds for $f$. And suppose $f$ is not uniformly contininuous. This means that (using logical negation):
$$\exists \varepsilon > 0: \forall \delta > 0: \lnot\left(\forall x,x' \in X : d(x, x') < \delta \rightarrow \rho(f(x), f(x')) < \varepsilon\right)$$
which we can rewrite (because an implication is only false when the left side is true and the right side is false) as
$$\exists \varepsilon > 0: \forall \delta > 0: \left( \exists x,x' \in X: d(x,x') < \delta \land \rho(f(x), f(x') \ge \varepsilon \right)$$
So we have this fixed $\varepsilon$ and we can pick arbitrarily close points in $X$ that have images always at least $\varepsilon$ apart. In particular we can pick points $x_n, z_n$ with this property and such that $d(x_n, z_n) < \frac{1}{n}$. Now the $(x_n), (z_n)$ have the property needed for the sequence property (their distances converge to $0$), but the sequences of their images are always $\varepsilon$ apart, so definitely their distances in $Y$ do not converge to $0$. This contradicts the sequence condition, so $f$ must be uniformly continuous.
