To find a counterexample in metric space. Suppose $X$ is a metric space, $z \in X$ and $(x_n)$ is a sequence in $X$. Show that if $X$ has a subsequence that converges to $z$, then dist$(z ,$ {$x_n :n ∈ N$}) $= 0$, and show also that the converse need not be true.
I have proved the first part that if $X$ has a subsequence that converges to $z$, then dist$(z ,$ {$x_n :n ∈ N$}) $= 0$. 
But I need help in finding the counterexample to show that the converse is not true.
Thank You!!

What about taking $X:=[0,1]$, $z:=1$ and
$$
x_n:=\begin{cases}1&\text{if }n=1,\\0&\text{otherwise}.\end{cases}
$$
