[proofplan] We discard finitely many initial terms and study the monotone tail beginning at the index where monotonicity starts. Since boundedness passes from the whole sequence to every tail, the tail is a bounded monotone real sequence. We prove directly from the least upper bound property of $\mathbb{R}$ that such a tail converges, using a supremum in the nondecreasing case and an infimum in the nonincreasing case. Finally, the $\varepsilon$-definition of convergence shows that convergence of the tail is exactly convergence of the original sequence. [/proofplan]

[step:Choose a monotone bounded tail] By eventual monotonicity, choose $N \in \mathbb{N}$ such that the tail $(a_n)_{n=N}^{\infty}$ is either nondecreasing or nonincreasing. Define the tail sequence $b: \mathbb{N} \to \mathbb{R}$ by $b_k = a_{N+k-1}$. Since $(a_n)$ is bounded, there exists $M \in [0,\infty)$ such that $|a_n| \le M$ for every $n \in \mathbb{N}$. Hence $|b_k| \le M$ for every $k \in \mathbb{N}$, so $(b_k)$ is bounded. Also, $(b_k)$ is nondecreasing if $(a_n)_{n=N}^{\infty}$ is nondecreasing, and nonincreasing if $(a_n)_{n=N}^{\infty}$ is nonincreasing. [/step]

[step:Prove that the monotone tail converges]We prove that $(b_k)$ converges. First suppose $(b_k)$ is nondecreasing. Define the set $B \subset \mathbb{R}$ by $B = \{b_k : k \in \mathbb{N}\}$. The set $B$ is nonempty and bounded above, so by the least upper bound property of $\mathbb{R}$ it has a supremum; define $L = \sup B$. Let $\varepsilon > 0$. Since $L-\varepsilon$ is not an upper bound for $B$, there exists $K \in \mathbb{N}$ such that $b_K > L-\varepsilon$. For every $k \ge K$, monotonicity gives $b_k \ge b_K$, while the definition of $L$ as an upper bound gives $b_k \le L$. Therefore $L-\varepsilon < b_k \le L$. Thus $|b_k-L| < \varepsilon$ for every $k \ge K$, so $b_k \to L$. Now suppose $(b_k)$ is nonincreasing. Define $C = \{b_k : k \in \mathbb{N}\}$. The set $C$ is nonempty and bounded below, so it has an infimum; define $L = \inf C$. Let $\varepsilon > 0$. Since $L+\varepsilon$ is not a lower bound for $C$, there exists $K \in \mathbb{N}$ such that $b_K < L+\varepsilon$. For every $k \ge K$, monotonicity gives $b_k \le b_K$, while the definition of $L$ as a lower bound gives $L \le b_k$. Therefore $L \le b_k < L+\varepsilon$. Thus $|b_k-L| < \varepsilon$ for every $k \ge K$, so $b_k \to L$.[/step]

[guided]We need to show that a bounded monotone real sequence has a limit. The proof uses the order completeness of $\mathbb{R}$: nonempty sets bounded above have suprema, and nonempty sets bounded below have infima. First assume that $(b_k)$ is nondecreasing. Define $B = \{b_k : k \in \mathbb{N}\}$. The set $B$ is nonempty because $b_1 \in B$. It is bounded above because the preceding step gave $|b_k| \le M$ for every $k \in \mathbb{N}$, hence $b_k \le M$ for every $k$. Therefore the least upper bound property gives a real number $L = \sup B$. We claim that $b_k \to L$. Let $\varepsilon > 0$. Since $L$ is the least upper bound of $B$, the number $L-\varepsilon$ cannot be an upper bound for $B$; otherwise $L-\varepsilon$ would be a smaller upper bound than $L$. Hence there exists $K \in \mathbb{N}$ such that $b_K > L-\varepsilon$. If $k \ge K$, then the nondecreasing property gives $b_k \ge b_K$. Since $L$ is an upper bound for $B$, we also have $b_k \le L$. Combining these inequalities gives $L-\varepsilon < b_k \le L$. This is exactly the estimate $|b_k-L| < \varepsilon$ for all $k \ge K$. Therefore $b_k \to L$. Now assume that $(b_k)$ is nonincreasing. Define $C = \{b_k : k \in \mathbb{N}\}$. The set $C$ is nonempty and bounded below because $|b_k| \le M$ implies $-M \le b_k$ for every $k$. Hence the greatest lower bound property gives a real number $L = \inf C$. Let $\varepsilon > 0$. Since $L+\varepsilon$ cannot be a lower bound for $C$, there exists $K \in \mathbb{N}$ such that $b_K < L+\varepsilon$. For every $k \ge K$, the nonincreasing property gives $b_k \le b_K$, and the definition of $L$ as a lower bound gives $L \le b_k$. Therefore $L \le b_k < L+\varepsilon$. Again this is exactly $|b_k-L| < \varepsilon$ for every $k \ge K$. Thus $b_k \to L$ in the nonincreasing case as well.[/guided]

[step:Transfer convergence from the tail to the original sequence] From the previous step, there exists $L \in \mathbb{R}$ such that $b_k \to L$. Let $\varepsilon > 0$. Choose $K \in \mathbb{N}$ such that $|b_k-L| < \varepsilon$ whenever $k \ge K$. Define $N_0 = N+K-1$. If $n \ge N_0$, then $k = n-N+1$ is an element of $\mathbb{N}$ satisfying $k \ge K$, and \begin{align*} a_n = b_{n-N+1}. \end{align*} Therefore $|a_n-L| < \varepsilon$ for every $n \ge N_0$. This proves that $a_n \to L$, so $(a_n)_{n=1}^{\infty}$ converges. [/step]