On a topology preliminary exam, students in past years were asked to find an immersion $f:S^1\to\Bbb R^2$ which cannot be approximated by an embedding (in the sense of the weak Whitney theorems).
I have an idea that seems reasonable: Trace out the Hawaiian earring, where the first $n$ circles are traversed between $0\leq\theta\leq 1-\frac1{2^n}$. This is a continuous map, and when we glue the pieces together at the origin we can probably do some partition of unity trick to avoid the abrupt change in magnitude of $f'$ that would otherwise occur when we changed pieces.
However, I can't figure out why there is no approximation by an embedding. I have some intuition: you have to end up on "one side" of a neighborhood of the origin, but you have to get back to the origin by $t=1$. This forces you to eventually leave the double of the neighborhood, but eventually the immersion says entirely inside the neighborhood. That seems problematic, although I'm not totally sure that those "eventually"s coincide enough to get an honest contradiction.
I'm interested in any answer to the question, but would be especially happy if you could talk about how to complete this argument, if it is true :P
Start with an embedding as an infinity shape where the immersed circle crosses itself once. Take a round circle $C$ centered on the intersection which intersects the circle in 4 points. There are two intersecting subarcs $\eta_1,\eta_2$ supported on the interior of the circle.
Now suppose there is a nearby embedding $E$ of the immersed circle. It may cross the circle $C$ in messy intersections but one can look at the set of open subarcs of $E$ supported on the interior of $C$. Each of arcs $\eta_i$ is close to one of these open subarcs. We will call their closures $\tilde{\eta}_i$. Now note that because the endpoints of $\tilde{\eta}_i$ are close to those of $\eta_i$, their endpoints are still combinatorially linked. Thus these arcs embedded on the interior of $C$ must actually cross. (You can use the Jordan curve theorem to prove this!) This contradicts that we had an embedding.