Cauchy Sequence: Multiplication Property This is part of my homework, but I could not solve it.

If $\{x_n\in\mathbb{R}\}$ and $\{y_n\in\mathbb{R}\}$ are both Cauchy sequences then $\{x_ny_n\}$
  is Cauchy sequence.

It is easy if I use the fact that $\{t_n\}$ is a Cauchy sequence if and only if it converges. But without using that fact, how can one solve it?
Edited.
 A: Try using the inequality 
$$|x_ny_n-x_my_m|=|x_ny_n-x_ny_m+x_ny_m-x_my_m|\leq|x_n||y_n-y_m|+|x_n-x_m||y_m|,$$ and note that you will want to use the fact that Cauchy sequences are bounded. 
A: Suppose we have an $\epsilon > 0$. We know for any $\epsilon_x > 0, \exists N_x$ s.t. $\forall m,n > N_x, |x_n - x_m| < \epsilon_x$. Similarly, for any $\epsilon_y > 0, \exists N_y$ s.t. $\forall m,n > N_y, |y_n - y_m| < \epsilon_y$.
Now consider $$|x_ny_n - x_my_m| = |x_ny_n - x_ny_m + x_ny_m - x_my_m|$$
$$\leq |x_n||y_n - y_m| + |y_m||x_n - x_m|$$
So if you choose $\epsilon_x < \frac{\epsilon}{2Y}$ and $\epsilon_y < \frac{\epsilon}{2X}$, where $X,Y$ are some upper bounds for the tails of the Cauchy sequences, then choosing the larger of $N_x$ and $N_y$ should suffice.
A: try this$$|x_ny_n-x_my_m|\le|x_ny_n-x_my_n|+|x_my_n-x_my_m|$$now when taking the common factor you'll have$$|x_ny_n-x_my_m|\le|y_n||x_n-x_m|+|x_m||y_n-y_m|$$now since $x_n$ and $y_n$ are cauchy that means they are bounded which implies that LHS goes to zero as $m,n\to\infty$ and hence $x_ny_n$ is cauchy.
