Bolzano-Weierstrass theorem Every sequence $\{ x_n \}_{n=1}^\infty$ bounded in $\mathbb R$ has, at least, a convergent subsequence.
Proof Since $\{ x_n \}_{n=1}^\infty$ is bounded, then $\exists a,b\in \mathbb R$, $a< b$, so $\{ x_n \}_{n=1}^\infty \subset [a,b]$.
If $\{x_n\}_{n=1}^\infty$ is a finite set, then it has a constant subsequence, which is convergent.
If $\{x_n\}_{n=1}^\infty$ is an infinite set, because of the Bolzano-Weierstrass theorem, there exists an accumulation point $x_0$ of $\{x_n\}_{n=1}^\infty$ (already proved).
Let's build a subsequence of $\{x_n\}_{n=1}^\infty$ that converges to $x_0$: since $\forall \varepsilon > 0 \;\, B(x_0, \varepsilon)\setminus\{x_0\}\cap \{x_n\}_{n=1}^\infty \neq \emptyset$ (because of the definition of accumulation point), let $\varepsilon_1 =1$, then $\exists n_1 \in \mathbb{N} $ s.t. $ x_{n_1} \in B(x_0, \varepsilon_1)$.
Let $m_2 \in \mathbb{N}$, $\varepsilon_2 = \frac{1}{m_2} < d(x_{n_1}, x_0)$, then $\exists n_2\in \mathbb{N} \;\, (n_2 > n_1)$ s.t. $ x_{n_2} \in B(x_0, \varepsilon_2)$.
Inductively we have that $\{x_{n_k}\}_{k=1}^\infty$ is a subsequence of $\{x_n\}_{n=1}^\infty$ and $n_k < n_{k+1}$, so $ d(x_{n_k}, x_0) < \varepsilon_k = \frac{1}{m_k}$.
Then $\lim_{k\to\infty} d(x_{n_k}, x_0) = 0$, which is equivalent to $\lim_{k\to\infty} x_{n_k}= x_0$ so $\{x_{n_k}\}_{k=1}^\infty$ is a convergent subsequence.
Question How can I rewrite the last part (the inductive part seems unclear to me) and everything unclear, since I can't understand why $\forall \varepsilon > 0 \;\, \exists k_0 \;\, \forall k \geq k_0 \;\, d(x_{n_k}, x_0) < \varepsilon $ (def. of convergence)...? Why is built $\{m_k\}$?