Limit of Subsequence of Bounded Sequence

Theorem
Let $$\left \langle {x_n} \right \rangle$$ be a sequence in $\mathbb{R}$.

Let $$\left \langle {x_n} \right \rangle$$ be bounded.

Let $$b \in \mathbb{R}$$ be a real number.

Suppose that $$\forall N: \exists n > N: x_n \ge b$$.

Then $$\left \langle {x_n} \right \rangle$$ has a subsequence which converges to a limit $$l \ge b$$.

Proof
Let us pick $$N \in \mathbb{N}$$.

Then $$\exists n_1 > N: x_{n_1} \ge b$$.

Again, $$\exists n_2 > n_1: x_{n_2} \ge b$$.

And so on: for each $$n_k$$ we find, $$\exists n_{k+1} > n_k: x_{n_{k+1}} \ge b$$.

In this way we can build a subsequence of $$\left \langle {x_n} \right \rangle$$ each of whose terms are $$b$$ or bigger.

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

Now, suppose $$x_{n_r} \to l$$ as $$r \to \infty$$.

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