Prove that $d(x_{m_i},x_{n_j})\geq c$ for all $i$ and $j$. Let $(x_k)$ be a divergent sequence in a metric space and suppose it has a convergent subsequence. Prove that there are two subsequences $(x_{m_i})^\infty _{i=1}$ and $(x_{n_j})^\infty _{j=1}$, and a positive real number $c$, such that $d(x_{m_i},x_{n_j})\geq c$ for all $i$ and $j$.
Well, since $(x_k)$ diverges two theorems come to mind: (1) that all subsequences of a convergent series converge to the same number and (2) all convergent sequences are Cauchy. I am not sure how the fact that $(x_k)$ has a convergent subsequence comes into play. Do I need to suppose that $(x_{m_i})^\infty _{i=1}$ and $(x_{n_j})^\infty _{j=1}$ are subsequences of the convergent subsequence and go from there? But doing that would require me to assume that $(x_k)$ has another convergent subsequence in order to prove anything useful, wouldn't it? 
$Proof.$ Let $(x_k)$ be a divergent sequence in a metric space and suppose it has a convergent subsequence which converges to $A$. Then there exists $\epsilon > 0$ such that for all $N>0$ there exists $n\geq N$ such that $|x_k - A| \geq \epsilon$.
Since $(x_k)$ is not convergent, it is not Cauchy. Then Then there exists $\epsilon > 0$ such that for all $N>0$ there exists $n, m\geq N$ such that $|x_n - x_m| \geq \epsilon$.
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Ok looking at what I've written I can see how this might turn out to be a 5 line proof, but I need help getting over the hump, because I can't see how to connect all the dots. 
 A: Hint Show that:


*

*If $(x_n)$ doesn't converge to $x$, then there exists a neighborhood $U$ of $x$ such that infinitely many $x_n$'s aren't members of $U$.

*If $x$ is a partial limit of $(x_n)$ (i.e., there exists a subsequence converging to $x$) then for every neighborhood $V$ of $x$, infinitely many $x_n$'s are members of $V$.


Then it's a matter of gainfully picking $U,V$ in a metric space.
A: Hint: Let $\{y_i \}$ be a convergent subsequence of $x_i$ which converges to $y$. This uses the fact that is has a convergence subsequence.
Hint: Using the fact that $x_i$ diverges, show that there exists a real number $a$, such that there are infinitely many points that are not in $B_a (y)$. 
Hint: Let $\{x_{m_i} \}$ be a suitable truncation of the sequence $\{y_i\}$, which converges to $y$.
Hint: Pick $\{x_{n_j} \}$ to be the infinitely many points that are not in $B_a (y) $.
Show that $d( x_{m_i}, x_{n_j} ) \geq \frac{a}{1000}$.

Bonus question: Do we really need the condition that there is a convergent subsequence?
