I understand that this isn't the standard proof provided in most introductory Real Analysis texts, but would like to see if this proof is valid.
Suppose that $\lim x_n = x$ and $\lim y_n = y$, and we have previously proved that $\lim x_n y_n = xy$.
By the multiplicative inverse property of real numbers, $\left( y_n \times \frac{1}{y_n} = 1 \right) \Rightarrow \lim \left( y_n \times \frac{1}{y_n} \right) = \lim 1 = 1$
Since $\lim x_n y_n = xy$, then $\lim \left( y_n \times \frac{1}{y_n} \right) = \lim y_n \times \lim \frac{1}{y_n} = y \times \lim \frac{1}{y_n} = 1 \iff \lim \frac{1}{y_n} = \frac{1}{y}$
Of course, this proof is only valid for $y \neq 0$ but this assumption holds since we assumed that the sequence $\frac{1}{y_n}$ is defined.