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 $$\left[{a \,. \, . \, b}\right]$$.