$f$ 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!
 A: 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.
