If a sequence in Hilbert space isn't Cauchy, then it hasn't convergent subsequence In a Hilbert space, if we want to show that a sequence $\;(y_n)\;$ hasn't a convergent subsequence , we show that $\;(y_n)\;$ isn't Cauchy sequence. I see this in many proofs but I 'm not sure I totally understand it. I guess it's quite trivial what I am about to ask but why does Cauchy property imply the above? 
My approach: 
By definition, since $\;(y_n)\;$ isn't Cauchy sequence there are $\;m,n \in \mathbb N \;$ and $\;n_0 \equiv n_0(n,m)\;$ such that $\;\forall m,n \ge n_0\;$ and $\;m \neq n\;$ : $\; \vert \vert y_n - y_m \vert \vert \ge M\;$ for $\;M\;$ positive constant. If I consider $\;n=k_n\;$ and $\;m=k_{n+1}\;$ then it follows $\;(y_n)\;$ cannot have a convergent subsequence.
EDIT: By  $\;(y_n)\;$ hasn't a convergent subsequence I mean that any subsequence of $\;y_n\;$ isn't convergent. 
Is this right or I missed something? Any help would be valuable. 
Thanks in advnace!
 A: It is difficult to decipher exactly what you are asking. You should prove that if $(x_n)$ has no convergent subsequences, then $(x_n)$ is not Cauchy. However, in your title and in your attempt, you seem to have these flipped; that is, it seems you are trying to prove that if a sequence isn't Cauchy, then it has no convergent subsequence. The latter is false of course, as can be seen by $x_n = 0$ for $n$ even and $x_n = 1$ for $n$ odd; this sequence is not Cauchy but has convergent subsequences. 
The former is true. In a Hilbert space, a sequence is Cauchy iff it is convergent. If a sequence is convergent then every subsequence is also convergent with the same limit. Conversely, if no subsequence converges, then the sequence itself cannot converge and thus cannot be Cauchy.
A: Let $x_n$ be a sequence and $C>0$ be so that for all $m \neq n$ we have 
$$\| x_n -x_m \| \geq C$$
Then $x_n$ has no converging subsequence.
Proof Assume by contradiction that $x_n$ has a convergent subsequence $x_{k_n}$. Then $x_{k_n}$ is Cauchy.
Pick some $0 < \epsilon <C$. Then, there exists some $N$ so that for all $m,n >N$ we have 
$$\| x_{k_n} -x_{k_m} \| < \epsilon <C$$
But if $m \neq n$ we also have by assumption
$$\| x_{k_n} -x_{k_m} \| \geq C$$
Contradiction.
