If $A$ is a rotation matrix, then $||Ax||=||x||$. Attempt:
Let $$
A =
\begin{bmatrix} 
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}, 
x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
$$
Then,
$$Ax =  
\begin{bmatrix} 
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \cos\theta \cdot x_1  -\sin\theta\cdot x_2 \\ \sin\theta \cdot x_1  +\cos\theta\cdot x_2 \end{bmatrix} $$
$$
\begin{align}
||Ax|| &= \sqrt{(\cos\theta \cdot x_1  -\sin\theta\cdot x_2)^2 + (\sin\theta \cdot x_1  +\cos\theta\cdot x_2)^2} \\
&=\sqrt{(\cos^2\theta \cdot x^2_1 -2\cos\theta \cdot x_1\sin\theta\cdot x_2+ \sin^2\theta\cdot x^2_2) + (\sin^2\theta \cdot x^2_1 +2\sin\theta \cdot x_1\cos\theta\cdot x_2+ \cos^2\theta\cdot x^2_2)} \\
&=\sqrt{x^2_1+x^2_2}
\end{align}
$$

However, my text seems to suggest that this is only true for $0\leq\theta\leq\pi$ here:

So where in my attempt did I go wrong?

What you have done is correct. The reason for putting the condition that $0\leq \theta \leq \pi$ is that the 'rotation of the plane' plays the role here. To make reader understand better, they have put the condition. For example , Let $\theta = 359^o$ then one might get confused whether rotation is by $1^o$ or $359^0$. To avoid this, they must have put the condition on $\theta$.
