Cauchy Sequence is Bounded

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Then every Cauchy sequence in $M$ is bounded.

Proof
Let $\left \langle {x_n} \right \rangle$ be a Cauchy sequence in $M$.

That is, $\forall \epsilon > 0: \exists N: \forall m, n > N: d \left({x_n, x_m}\right) < \epsilon$.

Particularly, setting $\epsilon = 1$, we have $\exists N_1: \forall m, n > N_1: d \left({x_n, x_m}\right) < 1$.

To show $\left \langle {x_n} \right \rangle$ is bounded, we need to show that there exists $a \in A$ and $K \in \R$ such that $d \left({x_n, a}\right) \le K$ for all $x_n \in \left \langle {x_n} \right \rangle$.

As $M = \left({A, d}\right)$ is a metric space, the Reverse Triangle Inequality holds: $\forall x, y, z \in A: \left|{d \left({x, z}\right) - d \left({y, z}\right)}\right| \le d \left({x, y}\right)$.

Now let $m = N_1 + 1$.

By the Reverse Triangle Inequality, for any $n > N_1$:

Now we take $K = \max \left\{{d \left({x_{N_1}, x_1}\right), d \left({x_{N_1}, x_2}\right), \ldots, d \left({x_{N_1}, x_{N_1}}\right), d \left({x_{N_1}, x_{N_1 + 1}}\right)}\right\} + 1$.

It follows that $\forall n \in \N^*: d \left({x_n, x_{N_1}}\right) \le K$ and so $\left \langle {x_n} \right \rangle$ is bounded in $M$.

This is clumsy - I tried to adapt it from the equivalent result for the real number line and it's not elegant. There must be a better way to do it.