Bolzano–Weierstrass theorem (bisection proof)

Let $(x_n)$ be bounded, so there exist $a<b$ with $x_n\in[a,b]$ for every $n$. We build a nested sequence of intervals by repeated bisection. Split $[a,b]$ into its two halves $[a,\tfrac{a+b}{2}]$ and $[\tfrac{a+b}{2},b]$. Since the sequence has infinitely many terms (counted with index) and only two halves are available, by the pigeonhole principle at least one half must contain $x_n$ for infinitely many indices $n$; call that half $[a_k,b_k]$ with $k=1$.

Repeat the same bisection on $[a_1,b_1]$: split it in two, and again by pigeonhole at least one half contains $x_n$ for infinitely many $n$; call it $[a_2,b_2]$. Continuing forever produces a nested chain $[a,b]\supset[a_1,b_1]\supset[a_2,b_2]\supset\cdots$, each containing $x_n$ for infinitely many indices, and each half the width of the previous one, so the width of $[a_k,b_k]$ is exactly $(b-a)/2^{k}$, which tends to $0$ as $k\to\infty$.

Now build the subsequence: since $[a_1,b_1]$ contains infinitely many terms of the sequence, pick any index $n_1$ with $x_{n_1}\in[a_1,b_1]$. Since $[a_2,b_2]$ also contains infinitely many terms (all but finitely many indices remain available), pick $n_2>n_1$ with $x_{n_2}\in[a_2,b_2]$. Continuing inductively, at each step $k$ choose $n_k>n_{k-1}$ with $x_{n_k}\in[a_k,b_k]$; this is always possible because $[a_k,b_k]$ contains infinitely many terms, so infinitely many indices beyond $n_{k-1}$ remain.

By the nested interval property of the real numbers (each $[a_k,b_k]$ is closed, nested, and their widths shrink to $0$), the intersection is a single point: $\bigcap_{k=1}^{\infty} [a_k,b_k] = \{L\}$ for some $L\in[a,b]$. Since both $x_{n_k}$ and $L$ lie in $[a_k,b_k]$, whose width is $(b-a)/2^{k}$, we get $|x_{n_k}-L|\le (b-a)/2^{k}\to 0$. As $k\to\infty$, the right-hand side tends to $0$, forcing $x_{n_k}\to L$. Thus $(x_{n_k})$ is a convergent subsequence of $(x_n)$, as required.