Squeeze Theorem/Sequences/Metric Spaces

Theorem
Let $(S,d)$ be a metric space.

Let $\langle x_n \rangle$ be a sequence in $S$.

Let $p \in S$.

Let $\langle r_n \rangle$ be a sequence in $\R_{\ge 0}$.

Suppose that $\langle r_n \rangle$ converges to $0$.

Suppose that for each $n$, $d(p,x_n) \le r_n$.

Then $\langle x_n \rangle$ converges to $p$.

Proof
Let $\epsilon > 0$.

Then for some $N$, $n > N \implies r_n < \epsilon$.

But then by Extended Transitivity, $n > N \implies d(p,x_n) < \epsilon$.

Thus $\langle x_n \rangle$ converges to $p$.