Solve the equation
$$
1+\frac{1}{1+\frac{1}{1+\ddots{ }_{1}+\frac{1}{x}}}=x \quad(n \text { layers }) .
$$
Solve the construction function
$$
f^{[n]}(x)=1+\frac{1}{1+\frac{1}{1+\ddots} 1+\frac{1}{x}} .
$$

Define
$$
f^{[0]}(x)=x, f^{[n]}(x)=1+\frac{1}{f^{[n-1]}(x)} .
$$
The fixed points of $f(x)$ are
$$
1+\frac{1}{x}=x \Rightarrow x_{1}=\frac{1+\sqrt{5}}{2}, x_{2}=\frac{1-\sqrt{5}}{2} .
$$

Since $M_{1} \subset M_{n}$, i.e., $\{x \mid f(x)=x\} \subset\left\{x \mid f^{[n]}(x)=x\right\}$, $\frac{1 \pm \sqrt{5}}{2}$ are the two roots of $f^{[n]}(x)=x$.
Next, prove that $f^{[n]}(x)=x$ has only two roots, i.e., $\left|M_{n}\right|=2$.
From
$$
1+\frac{1}{x}=\frac{x+1}{x}, 1+\frac{1}{\frac{x+1}{x}}=\frac{2 x+1}{x+1}, 1+\frac{1}{\frac{2 x+1}{x+1}}=\frac{3 x+2}{2 x+1}, \cdots
$$

we know that the original equation simplifies to $\frac{a x+b}{A x+B}=x$, which indicates that the equation has at most two real roots. Therefore, the roots of the equation can only be $\frac{1 \pm \sqrt{5}}{2}$.