Let $(a_n)$ be a bounded sequence in $\mathbb{R}$. Since $(a_n)$ is bounded, there exist $M_0, M_1 \in \mathbb{R}$ such that $a_n \in [M_0, M_1]$ for all $n$. We construct a convergent subsequence by repeatedly bisecting intervals.

Construction: Define a sequence of closed intervals $I_k = [a_k, b_k]$ and indices $n_1 < n_2 < n_3 < \cdots$ as follows:

• Base case ($k = 0$): Let $I_0 = [M_0, M_1]$. Since infinitely many terms of $(a_n)$ lie in $[M_0, M_1]$, choose $n_1$ such that $a_{n_1} \in I_0$.

• Inductive step: Suppose $I_k = [a_k, b_k]$ has been defined and contains infinitely many terms of $(a_n)$. Bisect $I_k$ into two halves:

  $\left[a_k, \frac{a_k + b_k}{2}\right] \quad \text{and} \quad \left[\frac{a_k + b_k}{2}, b_k\right].$

  At least one of these halves contains infinitely many terms of $(a_n)$. Choose such a half and call it $I_{k+1} = [a_{k+1}, b_{k+1}]$. Since $I_{k+1}$ contains infinitely many terms and we've already chosen $n_1, \ldots, n_k$, we can choose $n_{k+1} > n_k$ such that $a_{n_{k+1}} \in I_{k+1}$.

Properties of the construction:

1. $I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots$ (nested).
2. The length of $I_k$ is $|I_k| = (b_k - a_k) = (M_1 - M_0)/2^k \to 0$ as $k \to \infty$.
3. $a_{n_k} \in I_k$ for all $k$.

Convergence of the subsequence: By the nested intervals property (a consequence of completeness), there exists $L \in \mathbb{R}$ such that $L \in I_k$ for all $k$. (In fact, $\bigcap_{k=0}^\infty I_k = \{L\}$ since the intervals shrink to a point.)

Given $\epsilon > 0$, choose $K$ such that $|I_K| = (M_1 - M_0)/2^K < \epsilon$. Then for all $k \geq K$, both $a_{n_k}$ and $L$ lie in $I_k \subseteq I_K$, so

$|a_{n_k} - L| \leq |I_K| < \epsilon.$

Thus $a_{n_k} \to L$, completing the proof.

Let $(a_n)$ be bounded. Define

$L = \limsup_{n \to \infty} a_n = \lim_{k \to \infty} \sup_{n \geq k} a_n.$

Then $L$ is finite (since $(a_n)$ is bounded). For each $k$, there exists $n_k \geq k$ such that

$a_{n_k} > L - \frac{1}{k}$

(otherwise $L - 1/k$ would be an upper bound for $\{a_n \mid n \geq k\}$, contradicting the definition of $\sup$). Moreover, $a_{n_k} \leq \sup_{n \geq k} a_n$, so

$L - \frac{1}{k} < a_{n_k} \leq \sup_{n \geq k} a_n \to L.$

By the squeeze theorem, $a_{n_k} \to L$.