So we are having a matrix $A$:
$$\begin{bmatrix} 2&0&0\\ 0&1&1\\ 0&1&1\\ \end{bmatrix}$$
I already found the eigenvalues which are $\lambda_{1}=2$ (of multiplicity $2$) and $\lambda_2=0$
Then I found the eigenvectors which are $$\begin{bmatrix} 0\\ x_3\\ x_3\\ \end{bmatrix}$$
And
$$\begin{bmatrix} 0\\ -x_3\\ x_3\\ \end{bmatrix}$$
Here I'm stuck. As far as I know, because we have $3\times 3$ matrix we need $3$ vectors, but here I have only $2$, which means that our matrix is not diagonalizable. That's why I think I have a mistake somewhere.
The initial queation is to find a diagonalizaing matrix $S$, and a diagonal matrix $D$ such that
$$S^{-1}AS=D$$; $$A=SDS^{-1}$$
To find eigenvectors for the multiple eigenvalue we need to solve the homogeneous linear system
$$(2I-A)\vec x=\vec0\implies\;\text{the system is}\;\; y-z=0\implies y=z\implies \left\{\;\begin{pmatrix}0\\1\\1\end{pmatrix}\;,\;\;\begin{pmatrix}1\\0\\0\end{pmatrix}\;\right\}$$
is a basis for the eigenspace $\;V_2\;$ , so you must have made a mistake when evaluating this eigenspace