First, the definition. Let $(a_n)$ be a sequence in $\R$, we say $(a_n)$ is Cauchy, if for any $\epsilon> 0 $, we have $N>0$, such that for all $n,m>N$, we have $|a_n - a_m | < \epsilon$.
Lemma : Convergent sequence is Cauchy.
Pf: suppose $a_n \to a$. We need to show that for any $\epsilon> 0 $, we have $N>0$, such that for all $n,m>N$, we have $|a_n - a_m | < \epsilon$. Let $\epsilon_1 = \epsilon /2 $, then by convergence of $a_n$, we have $N_1>0$, such that $|a_n - a| < \epsilon_1$ for all $n > N_1$. Thus for any $n,m > N_1$, we have
$$ |a_n - a_m| \leq |a_n -a | + |a_m - a| \leq \epsilon_1 + \epsilon_1 = \epsilon $$
Thus, let $N=N_1$ would work.
Lemma : If $A_n \geq a_n \geq B_n$, and $\lim A_n = \lim B_n = a$, then $\lim a_n = a$.
Pf: for any $\epsilon > 0$, we have $N>0$ such that, for all $n > N$, $|A_n - a| < \epsilon, |B_n - a| < \epsilon$, then
$$ a_n \leq A_n \leq a+\epsilon, \quad a_n \geq B_n \geq a-\epsilon $$
Thus ,$|a_n - a| \leq \epsilon$ for all $ n > N$, hence $a_n \to a$.
Lemma : Let $(a_n)$ be a bounded sequence. $(a_n)$ is convergent if and only if $\limsup a_n = \liminf a_n$.
Pf: Assume $\limsup a_n = \liminf a_n = a$. Let $A_n = \sup_{m \geq n} a_m, B_n = \inf_{m \geq n} a_m$, then $A_n \geq a_n \geq B_n$. By preview lemma, we know $a_n \to a$.
Assume $(a_n)$ is convergent to $a$. Then, for any $\epsilon > 0$, there is an $N>0$, that for all $n > N$, $|a_n - a|<\epsilon$. In particular, we know $\limsup a_n \leq a+\epsilon$. Since this is true for all $\epsilon>0$, we have $\limsup a_n \leq a$. Similarly, $\liminf a_n \geq a$. On the other hand, since $A_n \geq B_n$, we have $\limsup a_n = \lim A_n \geq \lim B_n = \liminf a_n$. Hence, $\limsup a_n = \liminf a_n = a$.
Lemma: Cauchy sequence is bounded. (Exercise)
Lemma: If $\limsup a_n = A$, then for any $\epsilon > 0$, and any $N > 0$, there exists $n>N$ with $|a_n - A| < \epsilon$. Similarly, if $\limsup a_n = B$, then for any $\epsilon > 0$, and any $N > 0$, there exists $n>N$ with $|a_n - B| < \epsilon$.
Pf: Since $A_n = \sup_{m>n} a_m$ is a monotone decreasing sequence with limit $A$, then for any $\epsilon>0$ and $N>0$, we can find $M > N$ such that $A+\epsilon/2 > A_{M} \geq A$. By definition of $A_{M}$, there exists some $a_n$ with $n \geq M$, that $A_{M} \geq a_n > A_{M} - \epsilon/2$. Now, we have
$$ |a_n - A| \leq |a_n - A_M| + |A_M - A| \leq \epsilon/2 + \epsilon/2 = \epsilon $$
The result is proven.
Theorem : Cauchy sequence in $\R$ is convergent.
Proof: Let $(a_n)$ be a Cauchy sequence. By previous lemma, it is bounded. Let $A = \limsup a_n$ and $B = \liminf a_n$, we only need to show that $A = B$ to show $\lim a_n$ exists. We know $A \geq B$ for all bounded sequence $a_n$, suppose $A > B$, and let $\epsilon = (A-B)/3$. Then, by Cauchyness of $(a_n)$, we there is an $N>0$, such that for all $n,m> N$, we have $|a_n - a_m| < \epsilon$. By previous lemma, there exists $n>N$, with $|a_n - A|<\epsilon$, and $m>N$ with $|a_m - B| < \epsilon$. Hence
$$ |A - B| \leq |A-a_n|+|a_n - a_m| + |a_m - B| < 3 \epsilon = |A-B|, $$
notice the inequality is strict, hence we have a contradiction. Thus $A = B$.
If $n_1 < n_2 < \cdots $ is a strictly increasing sequence in $\N$, and $(a_n)$ is a sequence, then $(a_{n_k})$ is a subsequence of $(a_n)$.
In this section, we can ask, even if $(a_n)$ itself does not converge to some $a \in \R$, can we cherry-picking some nice subsequence in $(a_n)$ that does converge.
Example:
Definition: Let $(a_n)$ be a sequence in $\R$, if $a \in \R$ is the limit of a sequence of $(a_n)$, we say $a$ is a subsequential limit.
Lemma : $a$ is a subsequential limit of $(a_n)$ $\Leftrightarrow$ for any $\epsilon>0, N>0$, there exists an $n>N$, with $|a_n - a| < \epsilon$. Equivalently, for any $\epsilon>0$, the set $A_\epsilon = \{n \mid |a_n - a| < \epsilon \}$ is infinite.
Pf: $\Rightarrow$ : let $a_{n_k}$ be a subseq that converges to $n$, then we can find among member within this subsequence.
$ \Lefgarrow$: it's a good opportunity to introduce the Cantor's diagonal trick. For any positive integer $k$, we know $A_{1/k}$ is infinite, we write it in the $k$-th row, as $$ A_{1/k} = n_{k,1} < n_{k,2}< \cdots $$ As $k=1,2,\cdots$, we have an semi-infinite matrix of indices. We may check that $n_{k, i } \leq n_{k+1, i}$. Thus $n_{k,k} \leq n_{k+1, k} < n_{k+1, k+1}$, hence along the diagonal, we have strictly increasing sequence. Consider the subsequence $a_{n_{kk}}$, that will converge to $a$.