A sequence doesn't converges $\iff\exists\epsilon _0 >0$ and a subsequence $(x_{n_k})$ of $(x_n)$ such that $d(x_{n_k},x^*) \geq \epsilon _0$ Prove that sequence doesn't converges $\iff\exists\epsilon _0 >0$ and a subsequence $(x_{n_k})$ of $(x_n)$ such that $d(x_{n_k},x^*) \geq \epsilon _0$
My try: 
Definition of not convergence: there exists $\epsilon _0 >0$ such that $\forall N=N(\epsilon_ 0)\in \mathbb{N}$ exists $n\geq N$ such that $d(x_n,x^*)\geq \epsilon _0$. 


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*For $N=1$ exits $n_1 \geq 1$ such that $d(x_{n_1},x^*)\geq \epsilon _0$.

*For $N= n_1 +1, n_2 \geq N >n_1$ such that $d(x_{n_2},x^*)\geq \epsilon _0$

*If $n_1<n_2<...<n_k$ are such that $d(x_{n_j},x^*)\geq \epsilon _0$, we choose $N=n_k+1, n_k+1 \geq N$ such that $(d(x_{k+1},x^*) \geq \epsilon _0$.

*Finally, we obtain the subsequence $(x_{n_k})$ of $(x_n)$ such that $d(x_{n_k},x^*) \geq \epsilon _0$
I don't know if this is enough to prove both inclusions. Any suggestions would be great!

For the direction of $\Longleftarrow$:
Suppose on the contrary the sequence $(x_n)$ converges. Then for any $\epsilon>0$, there is some $M\in\mathbb{N}$ such that whenever $n\geq M$, one has $d(x_n,x)<\epsilon$. But by assumption, there is a subsequence $(x_{n_k})$ such that there exists $\epsilon_0>0$, $ N\in\mathbb{N}$ such that $d(x_{n_k},x)\geq\epsilon_0$ (for all $n_k\geq N$). Choose $\epsilon=\epsilon_0$. Then we get a contradiction. 
