Assuming that both $x_n$ and $y_n$ are cauchy I need to show that $x_ny_n$ is convergent. So I know I need to have two bounds
Pick an $M \gt 0$ and choose $N_1, N_2 \in \mathbb{N}$ such that $n \geq N_1$ implies $-\epsilon \lt x_n \lt \epsilon$ and $n \geq N_2$ implies $-\epsilon \lt y_n \lt \epsilon$ I am unsure of where to go from here
It seems that we may assume that $(x_n)_{n\geq1}$ and $(y_n)_{n\geq1}$ are Cauchy sequences in ${\mathbb R}$. This implies that there are numbers $\alpha$, $\beta\in{\mathbb R}$ with $\lim_{n\to\infty} x_n=\alpha, \>\lim_{n\to\infty} y_n=\beta$. Now $$x_ny_n-\alpha\beta=(x_n-\alpha)\beta+(y_n-\beta)\alpha+(x_n-\alpha)(y_n-\beta)\qquad(n\geq1)\ .$$ Given an $\epsilon>0$ there is an $N$ such that each of the three summands on the RHS here is $<{\epsilon\over3}$ for all $n>N$. Maybe you want to fill in the details for the summand $(x_n-\alpha)(y_n-\beta)$.