Convergent Subsequence in Closed Interval

Theorem
Let $\left[{a. . b}\right]$ be a closed real interval.

Then every sequence of points of $\left[{a. . b}\right]$ contains a subsequence which converges to a point in $\left[{a. . b}\right]$.

Proof
Let $\left \langle {x_n} \right \rangle$ be a sequence in $\left[{a. . b}\right]$.

Since $\left[{a. . b}\right]$ is bounded, so is $\left \langle {x_n} \right \rangle$.

By the Bolzano-Weierstrass Theorem, $\left \langle {x_n} \right \rangle$ has a subsequence $\left \langle {x_{n_r}} \right \rangle$ which is convergent.

Suppose $x_{n_r} \to l$ as $n \to \infty$.

Since $a \le x_{n_r} \le b$, from Lower and Upper Bounds for Sequences it follows that $a \le l \le b$.

So $\left \langle {x_{n_r}} \right \rangle$ converges to a point in $\left[{a. . b}\right]$.