algebraic geometry exercise: infinite subset is dense

A hypersurface $C \subset \mathbb{A}^2$ we call a plane curve. Show that any infinite subset of an irreducible plane curve $C \subset \mathbb{A}^2$ is dense in $C.$

Note. We call hypersurface the zero set $Z(f)$ of a non-constant polynomial $f \in k[x_1,\ldots,x_n],$ and we don't know the dimension of a hypersurface, nor that $\dim \mathbb{A}^n=n.$

Edit: $k$ algebraically closed.

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There is a quicker proof than mine given below, which is also more elementary. Namely, use the fact that two non-constant polynomials $f,g \in k[x,y]$ with no common factor have only finitely many common zeros (c.f. Fulton, "Algebaic Curves", p.9). – Nils Matthes Mar 26 '13 at 17:19
Thanks for the hints: using a corollary of your hint, which is proved in the book you mention, I can easily convince myself that $\dim\mathbb{A}^2=2;$ from this it is straightforward to solve the exercise. – jj_p Mar 27 '13 at 11:16

Hint: Every plane curve is homeomorphic (though not isomorphic!!) to $\mathbb{A}^1$ (why?). So it suffices to show that every infinite subset of $\mathbb{A}^1$ is dense in $\mathbb{A}^1$. This comes down (why?) to the fact that the only polynomial in $k[x]$ having infinitely many zeros is the zero polynomial.
Edit: I should add that I have $k$ algebraically closed in mind.