Bolzano-Weierstrass Theorem

Theorem
Every bounded sequence of real numbers has a convergent subsequence.

Proof (Monotonicity)
Let $\left \langle {x_n} \right \rangle$ be a bounded sequence in $\R$.

By the Peak Point Lemma, $\left \langle {x_n} \right \rangle$ has a monotone subsequence $\left \langle {x_{n_r}} \right \rangle$.

Since $\left \langle {x_n} \right \rangle$ is bounded, so is $\left \langle {x_{n_r}} \right \rangle$.

Hence, by the Monotone Convergence Theorem, the result follows.

Proof (Exhausting)
Let $(x_n)_{n \in \N}$ be a bounded sequence in $\R$.

By definition there are real numbers $c,C \in \R$ such that $c < x_n < C$. Then at least one of the sets $\{ x_n \mid c < x_n < \frac{c + C}{2}\}$, $\{ x_n \mid \frac{c + C}{2} < x_n < C \}$ and $\{ x_n \mid x_n =\frac{c + C}{2}\}$ contains infinitely many elements.

If the set $\{ x_n \mid x_n =\frac{c + C}{2}\}$ is infinite there's nothing to prove.

If this is not the case, choose the first element from the infinite set, say $x_{k_1}$.

Repeat this process for $(x_n)_{n > k_1}$.

As a result we obtain subsequence $(x_{k_n})_{n \in \N}$.

By construction $(x_{k_n})_{n \in \N}$ is a Cauchy sequence and therefore converges.