A bounded sequence with one accumulation point converges (rigorous proof) The proof of my question has been already given verbally here. I want to show it rigorously as follows.
1- Let $x^*$ be the accumulation point of $(x_k)$.
2- Suppose to the contrary that $(x_k) \not\to x^*$.
3- Then one can write the negation of the limit definition as follows:
$$
\exists\, \epsilon>0, \, \forall \, k_0 \, \exists \, k\geq k_0 \text{ and } \|x_k-x^*\| \geq \epsilon. \tag{1}
$$
4- $(x_k)$ is a bounded sequence therefore it has a subsequence $(x_{k_j})$ converges to $y^*$.
5- (Confusion) using (1) one should be able to write the following:
$$
\|x_{k_j}-x^*\| \geq \epsilon \, \forall \, k_j. \tag{2}
$$
6- Let $k_j \to \infty$ we get the following:
$$
\|y^*-x^*\| \geq \epsilon >0 
$$
which means $y^* \neq x^*$ that contradicts our assumption!
My question:
How can I use (1) to get (2) because I have already fixed $\epsilon$ and I do not know how to relate $\forall k_0$ in (1) to have $\forall k_j$. Also, I do not know what is the minimum of $k_j$?
Note:
Please address my question and do not use other arguments for proving the statement.
 A: 1- Let $x^*$ be the accumulation point of $(x_k)_{k \geq 1}$.
2- Suppose to the contrary that $(x_k) \not\to x^*$.
3- Then one can write the negation of the limit definition as follows:
$$
\exists\, \epsilon>0, \, \forall \, k^* \, \exists \, k\geq k^* \text{ s.t. } \|x_k-x^*\| \geq \epsilon. \tag{1}
$$
4- Generate a subsequence whose elements all satisfy (1) following below procedure:
4-1- Let $k^*$ be an arbitrary index. Then,
$$
\exists\ k_1\geq k^* \text{ s.t. } \|x_{k_1}-x^*\| \geq \epsilon.
$$
4-2- Let $k^*=k_1+1$. Therefore,
$$
\exists\ k_2\geq k^* \text{ s.t. } \|x_{k_2}-x^*\| \geq \epsilon.
$$
4-3- Let $k^*=k_2+1$ and so on.
4-4- Then, one gets $(x_{k_1}, x_{k_2}, \dots)=(x_{k_j})$ as a subsequence of $(x_k)$ whose elements satisfy $\|x_{k_j}-x^*\| \geq \epsilon$ for all $k_j$.
5- $(x_k)$ is a bounded sequence so is its subsequence, i.e., $(x_{k_j})$.
6- Since $(x_{k_j})$ is a bounded subsequence it has a convergent subsequence $(x_{k_{j_i}})$ converging to $y^*$.
8- Let $k_{j_i} $ go to infinity to get the following:
$$
\|y^*-x^*\| \geq \epsilon >0
$$
8- The above implies $y^* \neq x^*$ which contradicts our assumption!
9- Hence, $(x_k) \to x^*$.
A: It follows from $(1)$ that the set $\{k\in\Bbb N\mid|x_k-x^*|\geqslant\varepsilon\}$ is an infinite set of natural numbers. Let $m_k$ be its $k$th element. Then $(x_{m_k})_{k\in\Bbb N}$ is a sequence of real numbers such that$$(\forall k\in\Bbb N):|x_{m_k}-x^*|\geqslant\varepsilon.$$But each bounded sequence has a convergent subsequence. Let $(x_{n_k})_{k\in\Bbb N}$ be such a sequence and let $y^*$ be its limit. Since $\lim_{k\to\infty}x_{n_k}=y^*$ and since$$(\forall k\in\Bbb N):|x_{n_k}-x^*|\geqslant\varepsilon,$$we have\begin{align}|y^*-x^*|&=\left|\lim_{k\to\infty}x_{n_k}-x^*\right|\\&=\lim_{k\to\infty}|x_{n_k}-x^*|\\&\geqslant\varepsilon.\end{align}
