Here is one approach (another being this).

Let $ (x_n) $ be a sequence from $ [a,b] $. Then it has an accumulation point $ c \in [a,b] $ :

Let’s call an interval $ [p,q] $ “good” if $ x_n \in [p,q] $ for infinitely many $ n $.

Break $ I_0 := [a,b] $ into two equal parts $ [a, \frac{a+b}{2}] $ and $[\frac{a+b}{2}, b] $. Atleast one of these must be good, call it $ I_1 $.
Break $ I_1 $ into two equal parts. Atleast one of the resulting parts must be good, call it $ I_2 $. And so on. This gives a sequence of good intervals $ I_0 = [a,b] \supseteq I_1 \supseteq I_2 \supseteq \ldots .$

Pick $ x_{n_0} \in I_0 $. Now we can pick $ n_1 > n_0 $ with $ x_{n_1} \in I_1 $. Now we can pick $ n_2 > n_1 $ with $ x_{n_2} \in I_2 $, and so on.

This gives a subsequence $ (x_{n_k}) $ where $ x_{n_j}, x_{n_{j+1}}, \ldots $ lie in interval $ I_j $ (having length $ \frac{b-a}{2^j} $).
The subsequence is Cauchy.

For any $ j, k \geq M $, both $ x_{n_j}, x_{n_k} $ lie in $ I_M $ so $ | x_{n_j} - x_{n_k} | \leq \frac{b-a}{2^M}. $
Now given any $ \epsilon > 0 $, picking $ M $ such that $ \frac{b-a}{2^M} $ is $ < \epsilon $ to begin with, would do.

So subsequence converges, to a limit $ c \in \mathbb{R} $. Also $ a \leq x_{n_k} \leq b $ ensures $ a \leq c \leq b $.