If the sum of the partial distances is convergent, prove the sequence is Cauchy.

I am trying to prove that, given a sequence $(x_n)$ such that

$$ \sum_{n=1}^{\infty} d(x_n, x_{n+1}) < \infty$$

then the sequence is a Cauchy sequence. So far this is what I have:

Let $S_n = \sum_{i=1}^{n} d(x_i, x_{i+1})$. Then $S_n$ converges and, therefore, is a Cauchy sequence. So, given $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that if $N < m< n$, then:

$$|S_m - S_n| =$$$$=\left| \sum_{i=1}^{m} d(x_i, x_{i+1}) - \sum_{i=1}^{n} d(x_i, x_{i+1}) \right|$$$$=\left| \sum_{i=n+1}^{m} d(x_i, x_{i+1}) \right| < \varepsilon$$

But every summand is positive, so we must have each $d(x_i, x_{i+1}) < \varepsilon$ for every $i = n+1, \dots, m$. Therefore:

$$d(x_{n+1}, x_m) \leq d(x_{n+1}, x_{n+2}) + \dots + d(x_{m-1}, x_{m}) \leq (m-n-2) \cdot \varepsilon$$

This quantity is as small as desired, so $(x_n)$ is a Cauchy sequence.

I was pretty confident about this proof, but the more I ponder about it, the more I think it is wrong, because given $\varepsilon$ we can always take n and m such that m-n compensates $\varepsilon$. Is this reasoning correct? And if so, am I in the right track?