Is there a nice open set proof that multiplication is continuous? For students in a first course in analysis or topology, proving that certain function are continuous can be very tricky.  However, some proofs which are difficult for students to prove using the $\epsilon-\delta$ definition of continuity are much easier to prove using the topological definition that the pre-image of every open set be open.  For example, it is much easier for students to prove that $f(x)=x^2$ is continuous using open sets rather than $\epsilon-\delta$. One particularly challenging proof is showing that multiplication $\cdot \colon \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ is continuous.  Is anyone aware of a slick way to prove multiplication is continuous using open sets?  Any attempt I make seems to  more or less be just as complex as the $\epsilon-\delta$ definition.  
 A: Or you could (i) show (e,g by picture) that topologically speaking open balls are equivalent to open squares,
 (ii) then use the fact that $f(x,y)= xy $ is monotone in each variable $x,y>0$  and
 (iii) use  the  multiplicative properties of $f$  to show e,g that if $.9 x_0<x<1.1 x_0$ (and likewise for $y$), 
then $ .9^2  f(x_0, y_0) <f(x,y)< 1.1^2 f(x_0, y_0)$
In other words, make use of the multiplicative structure of the reals as much as possible. This proves continuity in the open first quadrant. 
A: Not sure if you would be interested in this approach, but it is easy to prove with the sequence definition of continuity. Suppose $x_n \to x$ and $y_n \to y$, then
$$
x_ny_n = (x + (x_n-x))(y + (y_n-y))  = xy + x(y_n-y) + y(x_n-x) + (x_n-x)(y_n-y)
$$
so that by triangle inequality
$$
|x_ny_n - xy| \leq |x| |y_n-y| + |y||x_n-x| + |x_n-x||y_n-y|
$$
since $x_n \to x, y_n \to y$ all three terms on the right tend to zero.
A: As nullUser said, if we take $x_n\to x$ and $y_n\to y$, we need to show that $x_n y_n \to xy$. However, the triangle inequality part can be a little bit simpler. Notice that $|x_n y_n - xy| = |x_n y_n - xy_n + xy_n - xy|$
$\leq |x_n y_n - xy_n| + |xy_n - xy| = |y_n||x_n-x| + |x||y_n-y|$.
Now, the right hand side goes to zero per assumption, meaning the product also converges.
