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Bolzano Weierstrass theorem states that

Every bounded sequence $\mathbb{R}$ has a convergent subsequence.

There are various ways to prove this, but I would like to ask if the following proof is incorrect as no elementary textbook use it.

Suppose $(x_n)_{n\geq1}$ is a bounded sequence. Then $A=\{x_1,x_2,\dots\}$ is non-empty and bounded subset of $\mathbb{R}$, thus $\sup A=\alpha$ exists in $\mathbb{R}$. For each $k\in\mathbb{Z}^+$ choose an element from $A$, call $n_k$ such that $\alpha-\frac{1}{k}<x_{n_k}\leq\alpha<\alpha+\frac{1}{k}$. Then it is clear that $x_{n_k}\to\alpha$.

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One problem is that for this argument to work, you would need $n_{k+1} > n_k$ which is not always possible. Indeed, let $x_1 = 1$ and $x_n = 0$ for all $n \ge 2$. The supremum of this sequence is $1$, but for each $k$ in you're example, we would have $n_k = 1$ and this does not define a subsequence.

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For the two distinct sequences $\{x_{n}\},\{y_{n}\}$, $x_{2n}=\dfrac{1}{2n}$, $x_{2n-1}=\dfrac{1}{2n-1}$, $y_{2n}=\dfrac{1}{2n-1}$, $y_{2n-1}=\dfrac{1}{2n}$, they have the same range $A=\{1/n: n=1,2,...\}$.

The proof concerns a subset of the range which has an adherent point, but that subset cannot "recover" back to the sequence. In such an example, the distinct sequences $\{x_{n}\}$ and $\{y_{n}\}$ are not distinguishable in terms of their ranges because they have the same range.

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