Suppose $f: S^2 \rightarrow {\bf R}^2$ is continuous; let $A$ be the set of points $u \in S^2$ such that $f(u)-f(-u) \in {\bf R} \times \{0\}$ (where $-u$ denotes the antipode of $u$). Given $u,-u \in A$, must there exist a path in $A$ joining $u$ to $-u$?

The appealing argument presented at https://www.youtube.com/watch?v=csInNn6pfT4 (which I have also seen elsewhere), which purports to prove the existence of a $u \in S^2$ with $f(u)-f(-u) = (0,0)$, hinges on this unstated assumption.

Can anyone devise an $f$ for which $A$ is something like a topologist's sine curve?

ADDED LATER: I like Ilya Bogdanov's example, and I wonder whether we can improve it. Is there an $f$ such that *no* point in $A$ can be joined to its antipode by a continuous curve?