Limit of Subsequence equals Limit of Sequence

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

Let $$\lim_{n \to \infty} x_n = l$$.

Let $$\left \langle {x_{n_r}} \right \rangle$$ be a subsequence of $$\left \langle {x_n} \right \rangle$$.

Then $$\lim_{n \to \infty} x_{n_r} = l$$.

That is, the limit of a convergent sequence equals the limit of a subsequence of it.

Proof
Let $$\epsilon > 0$$.

Since $$\lim_{n \to \infty} x_n = l$$, it follows that $$\exists N: \forall n > N: \left|{x_n - l}\right| < \epsilon$$.

Now let $$R = N$$.

Then for any $$\forall r > R: n_r \ge r$$ from Strictly Increasing Sequence of Natural Numbers‎.

Thus $$n_r > N$$ and so $$\left|{x_n - l}\right| < \epsilon$$.

The result follows.