I understand your statement now. I will summarize my answers here.
As I mentioned in my comments, $A$ is a rotation matrix since it can be written in the "standard" form under the new basis. In other words, it rotate any vector about the axis that is in the direction of $u_1$.
Notice that this last statement "it rotate any vector about the axis that is in the direction of $u_1$" is independent of the basis.
Your statement "to prove that there's a rotation such that for an orthonormal basis, applying the rotation to vectors of that basis gives A1, A2 and A3" is also correct. Because it also says the columns of $A$ is an orthonormal basis, which is true. But I think this is kind of a confusing way to prove a matrix is a rotation. Also you'll have to construct another rotation to prove it. And in what form would you construct another rotation?
A more straightforward way, as above, is to prove that the matrix $A$ rotates vectors.
Edit:
This is with regard to the matrix $A$ represented in terms of basis $u_1, u_2, u_3$. We can write a linear transformation $T$ as a matrix in terms of any basis using the following way. Given basis $e_1, e_2, e_3$, the matrix in terms of this basis is calculated by
$$[Te_1\quad Te_2 \quad Te_3]$$
where $Te_i$ is the column coordinate vector after you apply $T$ to $e_i$ in terms of this basis.
Looking at the question in this way, we see that $Au_1=\begin{pmatrix}1\\0\\0\end{pmatrix}$, $Au_2=\begin{pmatrix}0\\ \cos\theta\\ \sin\theta\end{pmatrix}$, $Au_3=\begin{pmatrix}0\\ -\sin\theta\\ \cos\theta\end{pmatrix}$.
Then
$$\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos\theta & -\sin\theta \\
0 & \sin\theta & \cos\theta \\
\end{bmatrix}$$
is exactly the matrix $A$ in terms of the basis $u_1, u_2, u_3$.
The vector $(1,0)$ along the $x$ axis is rotated into $\frac1{\sqrt2}(1,1)$ in the first quadrant. Thus $A$ rotates counterclockwise, which is by convention associated with positive angles; it represents a rotation by $+\frac\pi4$.
In a wider sense, one might also say that $A$ rotates by an angle $\frac\pi4$ if strictly speaking it rotated by $-\frac\pi4$. The distinction between the two is only valid in $\mathbb R^2$; in $\mathbb R^3$ the rotation by $-\frac\pi4$ around an axis is the rotation by $\frac\pi4$ around the inverse axis. As we tend to think of rotations in three dimensions, this reduction to positive rotation angles is sometimes also applied in talking about $\mathbb R^2$.
Best Answer
Hint:
You have simply to prove that $$ \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} =\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$
and this is a simple consequence of the identity $$ \sin^2 \theta + \cos^2 \theta=1 $$