The Bolzano-Weierstrass Theorem
Theorem 18.1 (The Bolzano-Weierstrass Theorem). Every bounded real sequence has a convergent subsequence.
Proof. Let $\{ a_n \}$ be a bounded real sequence. Then there exists a closed interval $[ c, d ]$ such that $a_n \in [ c, d ], \forall n \in \mathbb{P}$.
Consider the two subinterval, $\left[ c, \frac{c+d}{2} \right], \left[ \frac{c+d}{2}, d \right]$. One of these intervals must contain infinitely many terms of $\{ a_n \}$. We denote this interval by $[ c_1, d_1 ]$.
We repeat this process to $[c_1, d_1]$. One of the subintervals $\left[ c_1, \frac{c_1 + d_1}{2} \right], \left[ \frac{c_1 + d_1}{2}, d_1 \right]$ must contain infinitely many terms of $\{ a_n \}$. We denote this interval by $[c_2, d_2]$. Continuing this process, we obtain the sequence of closed intervals such that $$[c_1, d_1] \supset [c_2, d_2] \supset \cdots, \\ d_k - c_k = \frac{d - c}{2^k}, \forall k \in \mathbb{P}$$ and each interval $[ c_k, d_k ]$ contains $a_n$ for infinitely many positive integer $n$.
Choose $a_{n_1} \in [c_1, d_1]$. Since $[c_2, d_2]$ contains infinitely many terms of $\{ a_n \}$, we can choose $n_2 > n_1$ such that $a_{n_2} \in [c_2, d_2]$. Continuing the process, we obtain elements $a_{n_1}, a_{n_2}, ...$ satisfying $a_{n_k} \in [c_k, d_k], \forall k \in \mathbb{P}$ and $n_1 < n_2 < \cdots$. Then $\{ a_{n_k} \}$ is a subsequence of $\{a_n\}$.
Note that $\{ c_k \}$ is increasing and $\{ d_k \}$ is decreasing, and both are bounded. Then $$\lim_{n \to \infty} c_k = L \quad \text{ and } \quad \lim_{n \to \infty} d_k = M$$ by Theorem 16.2 and $$L - M = \lim_{k \to \infty} (d_k - c_k) = \lim_{k \to \infty} \frac{d - c}{2^k} = \lim_{k \to \infty} (d-c) \left( \frac{1}{2} \right)^k = 0 \\ \Longrightarrow L = M$$ by Corollary 12.4, Theorem 12.6 and Theorem 16.3.
Since $c_k \leq a_{n_k} \leq d_k$ for all positive integer $k$, $$\lim_{k \to \infty} a_{n_k} = L$$ by Theorem 14.3. Thus a subseqence $\{ a_{n_k} \}$ of $\{ a_n \}$ is convergent. $\blacksquare$