Show that $f$ is uniformly continuous $\iff$ $\bar d(f(x_n),f(y_n)) \to 0$ for all sequences $\{x_n\}, \{y_n\}$ in $X$ with $d(x_n, y_n) \to 0$. 
Suppose $(X, d), (Y,\bar d)$ are metric spaces, $f:X \longrightarrow Y$. Show that $f$ is uniformly continuous $\iff$ $\bar d(f(x_n),f(y_n)) \to 0$ for all sequences $\{x_n\}, \{y_n\}$ in $X$ with $d(x_n, y_n) \to 0$.

If $f$ is uniformly continuous then it is continuous. So, if $\{x_n\}$ is sequence in $X$ with $x_n \to x$, then $f(x_n) \to f(x)$. Similarly for $y$. So, if we have $d(x_n,y_n) \to 0$, then we will have $x_n=y_n$ for large $n$. This will imply that as $n \to \infty$,  $\ \bar d(f(x_n), f(y_n)) \to 0$.
Am I on the right track?
 A: Unfortunately no, you are not on the right track. 
Suppose $X:=\Bbb R$, let $x_n:=n$ and $y_n:=n+\frac 1n$, then
$$
d(x_n,y_n)=\frac 1n
$$
which converges to $0$ as $n\to \infty$. However, there is no such $x$ that $x_n\to x$.
You need to use the full power of uniform continuity to prove the statement, continuity is not enough.
A: Just work with the definition of uniform continuity (you need it, as BigbearZzz's answer shows). I write $|x-y|$ and $|f(x)-f(y)|$ to simplify the notation.
Assume $f$ is uniformly continuous, and let $(x_n)$, $(y_n)$ be two sequences with $|x_n-y_n|\to0$ $(n\to\infty)$. Let an $\epsilon>0$ be given. Then there is a $\delta>0$ with $|f(x)-f(y)|<\epsilon$ as soon as $|x-y|<\delta$. Furthermore there is an $N$ such that $|x_n-y_n|<\delta$, hence $|f(x_n)-f(y_n)|<\epsilon$,  for all $n> N$. As $\epsilon>0$ was arbitrary this proves $\lim_{n\to\infty}|f(x_n)-f(y_n)|=0$.
For the converse, assume that $f$ is not uniformly continuous. Then there is an $\epsilon_0>0$ for which there is no suitable $\delta>0$, meaning that for any proposed $\delta>0$ we can find two points $x$, $y$ with $|x-y|<\delta$ and $|f(x)-f(y)|\geq\epsilon_0$. In particular  for each $n\geq1$ we can find two points $x_n$, $y_n$ with $|x_n-y_n|<{1\over n}$ and $|f(x_n)-f(y_n)|\geq\epsilon_0$. For the two sequences $(x_n)$ and $(y_n)$ so "constructed" we have $\lim_{n\to\infty}|x_n-y_n|=0$, but $\lim_{n\to\infty}|f(x_n)-f(y_n)|=0$ does not hold.
