Topologist's Sine Curve not a regular submanifold of $\mathbb{R^2}$? I am trying to work out the details of following example from page 101 of Tu's An Introduction to Manifolds:

Example 9.3. Let $\Gamma$ be the graph of the function $f(x) =
 \sin(1/x)$ on the interval $]0, 1[$, and let $S$ be the union of
  $\Gamma$ and the open interval $I=\{(0,y)\in \mathbb{R^2} |−1<y<1\}$.
The subset $S$ of $\mathbb{R^2}$ is not a regular submanifold for the
  following reason: if $p$ is in the interval $I$, then there is no
  adapted chart containing $p$, since any sufficiently small
  neighborhood $U$ of $p$ in $\mathbb{R^2}$ intersects $S$ in infinitely
  many components.

Using Tu's definitions, I need to show that given a neighborhood $U$ of $p$, there exists no homeomorphism $\phi$ from $U$ onto an open set of $V \subset\mathbb{R^2}$ with the property that $U \cap S$ is the pre-image with respect to $\phi$ of the $x$ or $y$ axes intersected with $V$. 
I am not sure where the fact that $U \cap S$ has infinitely many components comes into play. I tried to derive a contradiction using this fact; but even if restrict my attention to connected neighborhoods $U$ of $p$, the intersection of the connected set $\phi(U)$ with the $x$ or $y$ axes might have infinitely many components (I think), so there's no contradiction with the fact that homeomorphism preserves components.
I would appreciate any help!

One simple way to see that $S$ is not a regular submanifold around $p$ is that it is not locally Euclidean: while there exist open subsets of $\mathbb{R}$ with as many connected components as you like, there are no points like $p$ in $\mathbb{R}$, that is, with every neighborhood not connected.
So, in the sentence "since any sufficiently small neighborhood $U$ of $p$ in $\mathbb{R}^2$ intersects $S$ in infinitely many components" the important part (besides "many components") is "any neighborhood".
