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.