Smooth closed real plane curve intersecting itself at infinitely many points

Question

Can a smooth closed real plane curve intersect itself at infinitely many points? It seems intuitively obvious that the answer should be no, yet I have no idea how to prove this or construct a counter-example. Here by smooth I mean $C^1$. If the answer is no, to which $C^k$ do we have to move for this geometric condition to be satisfied?

Edit: Here is an attempt to formalize the above: Let $C$ be a closed curve and $P$ a point at its image. We say that $C$ intersects itself at P, if for all parametrizations $f: [a,b] \to C$ (which are of the same $C^k$ class as C), the equation $f(x)=P$ has at least two solutions in $[a,b]$. I think this would work for what I had in mind posing this question.

By the way, I have no idea if this is the same with the transversal intersection definition proposed below.

Answer 1

Here is a less trivial example. The function

$$f(x)=\begin{cases} 0&\quad \text{if} \quad x=0\\ x^p \sin(1/x) &\text{otherwise} \end{cases}$$

is as smooth as you want (making $p$ large) but intersects the zero line infinitly often for $x\in[0,1]$. From this function you can easily make a closed loop intersecting itself infinitely often this way.

Answer 2

How about $y=x^2 \sin \frac 1x$ and $y=0$ on $x \in [0,1]$ plus a smooth turnaround at each end?