I think the way you started out is great. We need to show the following:
Given $\epsilon > 0$, there exists $N$ such that $n, m \geq N$ implies $|x_{n} - x_{m} | < \epsilon$.
But, as you correctly noted, given $\epsilon > 0$, $\exists N$ such that $n \geq N$ implies $|y_{n}| < \epsilon$.
But we have the inequality that for all $m \geq n$, $|x_{m} - x_{n} | \leq |y_{n}|$ (actually, this inequality also holds for all $n$, and in particular, for all $n \geq N$), which means $|x_{m} - x_{n} | < \epsilon$ for all $m \geq n$ (and all $n \geq N$). In other words, for all $n, m \geq N$, $|x_{m} - x_{n}| \leq |y_{n}| < \epsilon$.
So, given $\epsilon > 0$, we found our $N$ (the $N$ that made $|y_{n}| < \epsilon$ if $n \geq N$) such that $n, m \geq N$ implies $|x_{n} - x_{m}| < \epsilon$, which is exactly the definition of the sequence being Cauchy.
Lemma: For $x, y\in{\mathbb R}$, one has
$$|x^{1/3} - y^{1/3}|\le 3 |x - y|^{1/3}$$
Proof
Suppose that $|x|\le |x-y|$. It implies $|y|\le |x| + |y-x|\le 2|x - y|$. Hence $|x^{1/3} - y^{1/3}|\le |x|^{1/3}+ |y|^{1/3}\le
(1 + 2^{1/3})|x-y|^{1/3}\le 3 |x-y|^{1/3}$. The same proof holds if
one supposes $|y|\le |x-y|$
Suppose now that $|x-y|< \min(|x|, |y|)$. It follows that $x$ and $y$ have the same sign. We may suppose that they are positive. The mean value theorem gives for a $z\in (x, y)$
$$|x^{1/3} - y^{1/3}|\le \frac{1}{3} z^{-2/3}|x - y|\le \frac{1}{3} |x - y|^{-2/3}|x - y|\le \frac{1}{3}|x - y|^{1/3}$$
$\square$
The result is now obvious because for $\epsilon>0$, one can obtain $|x_n^{1/3} - x_m^{1/3}|\le \epsilon$ by imposing $|x_n - x_m|\le\epsilon^3/27$
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