On a disk with some properties in 3-space. Is it an immersed one?

Question

Suppose a disk $D$ mapped to $\mathbb{R}^3$ by the map $f: D \rightarrow \mathbb{R}^3$ such that the boundary of $D$ is embedded in $\mathbb{R}^3$ under $f$ and does not intersect with the interior of $D$. Also suppose that the boundary of $D$ forms an unknot in $3$-space. Let $\ell$ be a proper arc in $D$ such that in $3$-space $f(\ell) $ has a neighborhood in $f(D)$ which is fully twisted. My question is can we say that $D$ is an immersed disk in $3$-space. In other words, can we prove the existence of self intersections of the image $f(D)$ in $3$-space.

Answer 1

Either I need some clarification or I have a counterexample. In the picture below I show an example of a disk $D$ with boundary the unknot that has an embedded arc $\ell$ with a fully twisted neighborhood ($D$ itself). Am I misunderstanding "fully twisted"?

If we ask that $\ell$ is instead a simple closed curve, I think that there will be self-intersections. An approach might be as follows. Suppose the tip $n$ of a (tiny) normal vector field of the neighborhood of $\ell$ along $\ell$ generates the first homology of $\mathbb R^3$. If $D$ is embedded, then $\ell$ bounds a disk and hence $n$ intersects $D$ at least once. But since $n$ was tiny, this must mean that $D$ intersects itself, contradiction.

If that is indeed what you were asking I will try to provide a less sketchy answer.