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.

Solution 1:

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, $(x_k) \to x^*$.