I hadn't seriously attempted this problem until just now (I saw it last month and got scared

). But I just solved it without too much trouble!

Let $\displaystyle K=\mathbb{R}(x)$, the field of rational functions in $\displaystyle x$. Consider $\displaystyle f,g$ as elements of $\displaystyle K[y]$ (which is a P.I.D.). Then it is easy to see that $\displaystyle f,g$ are relatively prime in $\displaystyle K[y]$. Hence we can find rational functions $\displaystyle u(x), v(x)$ such that $\displaystyle u(x)f(x,y)+v(x)f(x,y)=1$. Multiplying through by a common denominator, we find that there exist polynomials $\displaystyle p(x), q(x), d(x)$ such that $\displaystyle p(x)f(x,y)+q(x)f(x,y)=d(x)$. The right side of this equation can vanish for only a finite number of possible values of $\displaystyle x$, hence there are only a finite number of values of $\displaystyle x$ which are candidates for an intersection point of the curves $\displaystyle f(x,y)=0$ and $\displaystyle g(x,y)=0$. Reversing the roles of $\displaystyle x,y$, we arrive at the same conclusion for $\displaystyle y$; hence there are only a finite number of possible intersection points of the two curves.