Why $\sin\left( \frac 1 x \right) $ oscillates infinitely many times as $x \to 0$

I believe your approach is correct.

As a little bonus, here is my explanation, that is more intuitive. Because $\frac{1}{x}$ grows from 1 to infinity as x goes from 1 to 0, $\sin(\frac{1}{x})$ must oscillate as many times within the 0 to 1 range as a regular $\sin(x)$ function must oscillate between 1 and infinity, which is an infinite number of oscillations. $\sin(\frac{1}{x})$ must fit an infinite number of oscillations between 0 and 1, thus, it has to oscillate very rapidly.You can make this argument for any number other than 1, because there must also be an infinite number of oscillations between 0 and, say, 0.0000000001, or 0.0000000000000001, or something even smaller.

Your argument is correct, but you have a gap, some minor terminological issues, and the presentation is poor.

For the terminological issue, saying: "assume a period $p$ such that $\sin(1/x) = \sin(1/(x+p))$" strongly suggests that you are saying $\sin(1/x)$ is a periodic function with period $p$ which would mean $p$ is constant. Of course, as you show, $p$ is not constant, so this terminology is misleading.

The gap in your argument is when you say $\frac{1}{x}-\frac{1}{x+p} = K$ for a positive constant $K$. This can be made to be true, but you give no argument why, and the order of your presentation makes this not necessarily, let alone obviously, true. If, at this point in the argument, someone was still under the impression that $p$ was a constant, then this would be obviously false. Even if the reader hadn't assumed $p$ was constant, there's still no given reason to believe this is true.

The problem with your presentation (beyond the terminological one) is that you first assume that there is some function $p$ with the desired property, then you assume a statement which need not be true. It's not necessarily the case that if $\sin(1/x)=\sin(1/(x+p))$ that $\frac{1}{x}-\frac{1}{x+p}$ is a positive constant. You certainly give no argument for it. It's also not true for arbitrary $K$, even without knowing what $p$ is. You are not claiming this, but you give no argument any such $K$ exists. (From the perspective of first-order logic, essentially what's happening is that you are mixing around the order of quantifiers.)

Here's a rationalized version of what I think your intent was:

First, note that $\sin$ is a periodic function with period $2\pi k$ for any integer $k$. Set $K = 2\pi k$ for some positive $k$. Now $\sin(1/x) = \sin(1/x - K)$. Find $p(x)$ such that $\frac{1}{x} - K =\frac{1}{x+p(x)}$. We can do this by solving for $p(x)$ getting $p(x) = \frac{x}{1-Kx}-x$. From here proceed as the remainder of your argument.

Per MathTrain's and my comments on the question, whether this proves that $\sin(1/x)$ "oscillates rapidly" depends on what "oscillates rapidly" means.

You have shown that if there is a period, then it must be infinitesimally small, which implies that the function under consideration must be constant. However, you don't consider the case where there is no period, which is the case that this problem falls under. Since $\sin\dfrac{1}{x}$ is not constant, then what you proved implies that $\sin\dfrac{1}{x}$ is not periodic.

Another approach is to consider the zeroes of the function $\sin\dfrac{1}{x}$.

Note that $\sin\dfrac{1}{x}=0$ if and only if $x=\dfrac{1}{k\pi}$, for some $k\in\mathbb{N}$. The sequence $a_k = \dfrac{1}{k\pi}$ approaches $0$ from above as $k\rightarrow\infty$, so then $\forall \varepsilon > 0$ there exist infinitely many $x\in(0,\varepsilon]$ such that $\sin\dfrac{1}{x}=0$. Each place that $\sin\dfrac{1}{x}$ hits zero corresponds to one half of a cycle of sine, so there are infinitely many oscillations in $(0,\varepsilon]$.