Prove this number fact Prove that $x \neq 0,y \neq 0 \Rightarrow xy \neq 0$. 

Suppose $xy = 0$. Then $\frac{xy}{xy} = 1$. Can we say that $\frac{xy}{xy} = 0$ and hence $1 = 0$ which is a contradiction? I thought $\frac{0}{0}$ was undefined.
The proof is valid in any field $\rm\: K$ (though it might be circular depending on the context). Namely, $\rm\:0\ne x,y\in K\:$ $\:\Rightarrow\:$ $\rm\: 1/x,1/y\in K\:$ $\:\Rightarrow\:$ $\rm\:(1/x)(1/y) = 1/(xy)\in K\:$ $\:\Rightarrow\:$ $\rm\:xy\ne 0\:.$ The OP's proof is simply this proof recast into a proof by contradiction. To be precise the OP's proof is as follows:

As above, $\rm\: x,y\ne 0\ \Rightarrow\ z := 1/(xy)\in K\:,\:$ i.e. $\rm\:xyz = 1\:.\:$ So $\rm\ xy=0\ \Rightarrow\ 0 = 1\:,\:$ a contradiction.  

That's precisely the OP's proof, except I've replaced $\rm\:xy/(xy)\:$  by $\rm\:xyz\:$ to avoid possible confusion.

This is a valid proof. The confusion stems from the fact that it is a proof by contradiction. Such proofs - by their very nature - may encounter all sorts of strange looking mathematical objects, such as the above expression of the form $\rm\: 0/0 = 1\:.\:$ This is just $\rm\:1/1 = 1\:$ in the trivial ring $\:\{0\}\:$ where $\rm\:0 = 1\:.\:$ However, the trivial ring is not a field, since $\rm\:0\ne 1\:$ by the definition of a field (or integral domain). So, as above, $\rm\:0 = 1\:$ is a common target for proofs by contradiction in a field.

Proofs by contradiction often prove immensely confusing to students when first encountered. Learning to wrap one's mind around the bizarre contradictory objects encountered in such proofs is skill that comes with practice. A striking example of such confusion is Euclid's classical proof that there are infinitely many primes. Although Euclid's proof was constructive, it is widely presented as a proof by contradiction (and falsely claimed that this was Euclid's proof). When presented in contradictory form this proof often leads to much confusion. There are hundreds of threads on sci.math permeated by such confusion. One can reach all sorts of contradictions to terminate Euclid's proof, e.g. $\rm\:0 = 1\:$ or $\rm\: 1\:$ is prime, or some integer is both prime and composite, etc. Indeed, one can deduce anything in a contradictory theory such as the integers with finitely many primes. Such contradictions often prove too much to grasp for many beginners. Apparently this is because  we have such strong intuition about integers that one contradiction easily implies many others, and this quickly grows too much to handle intuitively. This does not occur to the same degree when one works with more abstract structures, where real-world intuition has less chance to restrain logical thought processes. Such is the strange nature of proofs by contradiction. 

Note $\ $ The OP has revealed the source as Proposition 1.16 in Rudin's Principles of Mathematial Analysis. I've appended it below. It is essentially as I surmised above.
consider that xy=0 from the  basic  axioms we know that  product is zero  when  at least one of the  multiplier is  zero  so is means that  x is zero or y is zero  or both x and y is equal zero  which contradicts  to given condition that neither x or y is zero so it means that their product is also  not equal to zero