Question about my construction of $x$ to prove: If $g(x)=x^2\sin(\frac{1}{x})$, then there are numbers $x$ arbitrarily close to $0$ with $g'(x)=1$ Chapter 11 Question 67b of Spivak's Calculus asks the reader to show the following:

If $g(x)=x^2\sin(\frac{1}{x})$, then there are numbers $x$ arbitrarily close to $0$ with $g'(x)=1$.

IMPORTANTLY:
I will walk through how I constructed the $x$...which will seem logically inconsistent, because I assume the answer and work backwards...the purpose of this post is to evaluate the method for generating the $x$, rather than to state the proof.

Firstly, $g'(x)=2x\sin(\frac{1}{x})-\cos(\frac{1}{x})$.
The goal of this poof is to show that for any $\delta \gt 0$, I can construct an $x \in (0,\delta)$ such that $g'(x)=1$.
The easiest approach to doing this is finding an $x$ such that $\sin(\frac{1}{x})=0$ and $\cos(\frac{1}{x})=-1$.
This occurs when $\frac{1}{x}=\pi + 2\pi n$ for some $n \in \mathbb N$.
Solving for $x$, we get that $x = \frac{1}{\pi(1+2n)}$ for some $n$. Next, we need to figure out how we generate this $n$.

Because $\mathbb N$ is unbounded, we know that for any $\varepsilon\gt 0$, there is an $n'$ such that $\frac{1}{n'}\lt\varepsilon\quad (\dagger_1)$.

$\color{red}{\text{Consider some arbitrary } \delta}$.
I need $0\lt x \lt \delta$ and I need $x=\frac{1}{\pi(1+2n)}$ for some not yet determined $n$, so I force the following statement: $0 \lt \frac{1}{\pi(1+2n)} \lt \delta$. Next, I am going to solve for $n$:
\begin{align}
&\frac{1}{\pi(1+2n)}\lt \delta \quad \implies \\
&1 \lt \delta\pi+2n\delta\pi \quad \implies \\
& 1-\delta\pi \lt 2n\delta\pi \quad \implies \\
&\frac{1-\delta\pi}{2\pi\delta}\lt n \quad(\dagger_2)
\end{align}
What I want to do at this point is employ the following lemma: $a \gt 0 \land b \gt 0 \land a \gt b \rightarrow \frac{1}{a}\lt\frac{1}{b} (*)$. However, the nuumerator of $(\dagger_2)$ is concerning because $1-\delta\pi \leq 0$ for certain values of $\delta$: in particular, when $\delta \geq \frac{1}{\pi}$.
Therefore, I am going to alter my original statement about "$\color{red}{\text{Consider some arbitrary } \delta}$" and change it to any $\delta \lt \frac{1}{\pi}$.
Then, we can apply our lemma $(*)$ to $(\dagger_2)$ to say:
$$\frac{1}{n} \lt \frac{2\pi\delta}{1-\delta\pi}$$
Therefore, if I can find an $n$ that satisfies this statement, I am done. Let $\frac{2\pi\delta}{1-\delta\pi}$ be one such instance of a $\varepsilon$ and apply $(\dagger_1)$. Then we know that such an $n$ exists.
At this point, all of our steps can proceed in reverse, and we show that for any $\delta \lt \frac{1}{\pi}$, we can find an $n$ such that there is an $x \in (0,\delta)$ where $g'(x)=0 \quad (\dagger_3)$.
To address the remaining $\delta \geq \frac{1}{\pi}$, simply choose some $x$ satisfying a $\delta' \lt \frac{1}{\pi}$. For example, for any $\delta \geq \frac{1}{\pi}$, choose the $x$ associated with $\delta'=\frac{1}{\pi+1}$. Given that $\frac{1}{\pi+1} \lt \frac{1}{\pi}$, we know that this exists because of $(\dagger_3)$.
$\square$

Is this the appropriate approach?

You can simplify $2x\sin(\frac{1}{x})-\cos(\frac{1}{x}) =1$ by using some double angle formulae for trigonometric functions.
\begin{array}
\, 2x\sin(\frac{1}{x})-\cos(\frac{1}{x}) =1 & \iff 2x\sin(\frac{1}{x}) = \cos(\frac{1}{x}) + 1 \\
& \iff 4x\sin(\frac{1}{2x}) \cos(\frac{1}{2x}) = 2 \cos^2 (\frac{1}{2x})\\
& \iff \tan (\frac{1}{2x}) = \frac{1}{2x}
\end{array}
By letting $u=\frac{1}{2x}$, the equation becomes $u = \tan u$. And, you should be able to see there are infinitely many solutions as $u \to \infty$.
