[proofplan] We first prove the theorem in $\mathbb{R}$ by repeatedly bisecting a bounded interval and retaining a half-interval containing infinitely many [sequence](/page/Sequence) terms. The nested closed intervals have diameters tending to zero, so their intersection is a single point, and the selected terms [converge](/page/Convergence%20(Real%20Sequences)) to that point. For $\mathbb{R}^n$, we apply the one-dimensional result successively to each coordinate, taking subsequences of subsequences until all coordinates converge, then use coordinate convergence to obtain convergence in Euclidean norm. [/proofplan] [step:Construct a convergent subsequence for bounded real sequences] Let $\{a_k\}_{k=1}^{\infty} \subset \mathbb{R}$ be a bounded real sequence. Choose [real numbers](/page/Real%20Numbers) $A,B \in \mathbb{R}$ such that $A \le a_k \le B$ for every $k \in \mathbb{N}$, and define the initial closed interval $I_0 := [A,B]$. We construct by induction a nested sequence of closed intervals $\{I_m\}_{m=0}^{\infty}$ and a strictly increasing sequence of indices $\{k_m\}_{m=1}^{\infty} \subset \mathbb{N}$ such that $I_m$ contains infinitely many terms of $\{a_k\}_{k=1}^{\infty}$, $a_{k_m} \in I_m$, and \begin{align*} \operatorname{diam}(I_m) = \frac{B-A}{2^m} \end{align*} for every $m \in \mathbb{N}$. Assume $I_{m-1} = [\alpha_{m-1},\beta_{m-1}]$ has been constructed and contains infinitely many sequence terms. Define its two closed halves by \begin{align*} I_{m,-} &:= \left[\alpha_{m-1}, \frac{\alpha_{m-1}+\beta_{m-1}}{2}\right],\\ I_{m,+} &:= \left[\frac{\alpha_{m-1}+\beta_{m-1}}{2}, \beta_{m-1}\right]. \end{align*} Since every term lying in $I_{m-1}$ lies in $I_{m,-} \cup I_{m,+}$, at least one of $I_{m,-}$ and $I_{m,+}$ contains infinitely many terms of the sequence. Choose one such half and call it $I_m$. Since $I_m$ contains infinitely many terms, choose $k_m > k_{m-1}$ with $a_{k_m} \in I_m$, where for $m=1$ there is no previous index restriction except $k_1 \in \mathbb{N}$. Thus $\{a_{k_m}\}_{m=1}^{\infty}$ is a subsequence of $\{a_k\}_{k=1}^{\infty}$, and the closed intervals satisfy \begin{align*} I_0 \supset I_1 \supset I_2 \supset \cdots, \qquad \operatorname{diam}(I_m) = \frac{B-A}{2^m}. \end{align*} By the [Cantor Intersection Theorem](/theorems/624), applied to this nested sequence of nonempty closed bounded intervals whose diameters tend to $0$, there exists a unique $L \in \mathbb{R}$ such that \begin{align*} \bigcap_{m=0}^{\infty} I_m = \{L\}. \end{align*} For each $m \in \mathbb{N}$, both $a_{k_m}$ and $L$ belong to $I_m$, hence \begin{align*} |a_{k_m}-L| \le \operatorname{diam}(I_m) = \frac{B-A}{2^m}. \end{align*} Since $(B-A)/2^m \to 0$ as $m \to \infty$, the squeeze property for real sequences gives $a_{k_m} \to L$. [guided] We begin with a bounded real sequence $\{a_k\}_{k=1}^{\infty} \subset \mathbb{R}$. Boundedness means that there are real numbers $A,B \in \mathbb{R}$ such that \begin{align*} A \le a_k \le B \end{align*} for every $k \in \mathbb{N}$. Define $I_0 := [A,B]$. The purpose of the bisection construction is to keep narrowing the interval while ensuring that infinitely many sequence terms remain available inside it. Suppose we have already chosen a closed interval $I_{m-1} = [\alpha_{m-1},\beta_{m-1}]$ containing infinitely many terms of the sequence. Split it into the two closed halves \begin{align*} I_{m,-} &:= \left[\alpha_{m-1}, \frac{\alpha_{m-1}+\beta_{m-1}}{2}\right],\\ I_{m,+} &:= \left[\frac{\alpha_{m-1}+\beta_{m-1}}{2}, \beta_{m-1}\right]. \end{align*} Every sequence term in $I_{m-1}$ belongs to $I_{m,-} \cup I_{m,+}$. If both halves contained only finitely many terms, then their union would contain only finitely many terms, contradicting the inductive assumption that $I_{m-1}$ contains infinitely many terms. Therefore at least one half contains infinitely many terms; choose such a half and call it $I_m$. Because $I_m$ contains infinitely many terms of the original sequence, there is an index larger than all previously chosen indices whose term lies in $I_m$. Thus we may choose $k_m > k_{m-1}$ with $a_{k_m} \in I_m$, with the evident modification at $m=1$. This produces a genuine subsequence $\{a_{k_m}\}_{m=1}^{\infty}$ because the indices are strictly increasing. The construction also gives nested closed intervals \begin{align*} I_0 \supset I_1 \supset I_2 \supset \cdots \end{align*} and each bisection halves the diameter, so \begin{align*} \operatorname{diam}(I_m) = \frac{B-A}{2^m}. \end{align*} The hypotheses of the Cantor Intersection Theorem are satisfied: the intervals are nonempty, closed, bounded, nested, and their diameters tend to $0$. Hence there is a unique point $L \in \mathbb{R}$ such that \begin{align*} \bigcap_{m=0}^{\infty} I_m = \{L\}. \end{align*} Finally, for every $m \in \mathbb{N}$, the selected term $a_{k_m}$ lies in $I_m$, and the point $L$ also lies in $I_m$. Two points in an interval are separated by at most the interval diameter, so \begin{align*} |a_{k_m}-L| \le \operatorname{diam}(I_m) = \frac{B-A}{2^m}. \end{align*} The right-hand side tends to $0$, so $a_{k_m} \to L$ as $m \to \infty$. This proves the one-dimensional Bolzano-Weierstrass theorem. [/guided] [/step] [step:Extract subsequences coordinate by coordinate in $\mathbb{R}^n$] Let $\{x_k\}_{k=1}^{\infty} \subset \mathbb{R}^n$ satisfy $\|x_k\| \le M$ for some $M > 0$ and every $k \in \mathbb{N}$. For each $k \in \mathbb{N}$, write \begin{align*} x_k = (x_{k,1},\dots,x_{k,n}). \end{align*} For every coordinate index $i \in \{1,\dots,n\}$ and every $k \in \mathbb{N}$, the Euclidean coordinate bound gives \begin{align*} |x_{k,i}| \le \|x_k\| \le M. \end{align*} Thus each coordinate sequence is bounded in $\mathbb{R}$. Define $y_{0,j} := x_j$ for $j \in \mathbb{N}$. Applying the one-dimensional result to the first coordinate sequence of $\{y_{0,j}\}_{j=1}^{\infty}$, choose a subsequence $\{y_{1,j}\}_{j=1}^{\infty}$ whose first coordinates converge. Next apply the one-dimensional result to the second coordinate sequence of $\{y_{1,j}\}_{j=1}^{\infty}$, and choose a subsequence $\{y_{2,j}\}_{j=1}^{\infty}$ whose second coordinates converge. Since a subsequence of a convergent real sequence converges to the same limit, the first coordinates of $\{y_{2,j}\}_{j=1}^{\infty}$ still converge. Continuing this procedure through coordinate $n$, we obtain a subsequence $\{y_{n,j}\}_{j=1}^{\infty}$ of the original sequence such that, for each $i \in \{1,\dots,n\}$, the real sequence of $i$-th coordinates of $\{y_{n,j}\}_{j=1}^{\infty}$ converges. Therefore there exist real numbers $L_1,\dots,L_n \in \mathbb{R}$ such that \begin{align*} (y_{n,j})_i \to L_i \end{align*} as $j \to \infty$ for every $i \in \{1,\dots,n\}$. [/step] [step:Convert coordinate convergence into convergence in $\mathbb{R}^n$] Define the point $x \in \mathbb{R}^n$ by \begin{align*} x := (L_1,\dots,L_n). \end{align*} Since $\{y_{n,j}\}_{j=1}^{\infty}$ is a subsequence of $\{x_k\}_{k=1}^{\infty}$, there is a strictly increasing sequence of indices $\{k_j\}_{j=1}^{\infty} \subset \mathbb{N}$ such that $y_{n,j} = x_{k_j}$ for every $j \in \mathbb{N}$. Let $\varepsilon > 0$. For each $i \in \{1,\dots,n\}$, coordinate convergence gives an index $N_i \in \mathbb{N}$ such that \begin{align*} |(y_{n,j})_i - L_i| < \frac{\varepsilon}{n} \end{align*} whenever $j \ge N_i$. Define $N := \max\{N_1,\dots,N_n\}$. Then for every $j \ge N$, the triangle inequality for the Euclidean norm gives \begin{align*} \|x_{k_j}-x\| &= \|y_{n,j}-x\| \\ &\le \sum_{i=1}^{n} |(y_{n,j})_i - L_i| \\ &< \sum_{i=1}^{n} \frac{\varepsilon}{n} \\ &= \varepsilon. \end{align*} Hence $x_{k_j} \to x$ as $j \to \infty$. This proves that every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence. [/step]