Show that this matrix is not diagonalizable
No, it's not diagonalizable. If the two eigenvalues of a $ 2 \times 2 $ matrix were distinct, it would be; when they're the same, it might be (but in this case it's not).
The eigenvalues of an $n \times n$ matrix turn out (as you'll probably learn soon) to be the roots of a degree-$n$ polynomial. Since every degree-$n$ polynomial has $n$ roots (when counted with multiplicity, and allowing for complex roots as well as real ones), this means that every $n \times n$ matrix has $n$ eigenvalues (when counted with algebraic multiplicities).
By the way, it appears that you've done exactly the right thing to determine how many eigenvectors there are that correspond to a given evalue; in general, there's no obvious and simple way to do it except to look for the solution space of an associated system of equations, as you did.
HINT
Recall what are the necessary and sufficient conditions for a matrix to be diagonalizable and note that here we have an eigenvalue $2$ with arithmetic multiplicity $2$ and geometric multiplicity $1$, that is an eigenspace with dimension $1$.
- No. $A - 2I$ has only one linearly independent column. (The second column of $A-2I$ is zero.)
- To give a quick example, let's consider the 2D rotation matrix. $$\begin{pmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{pmatrix}$$ Eigenspace for a linear transformation is an example of invariant subspace. Since there's no proper invariant subspace in 2D rotation, the rotation matrix doesn't have real eigenvalues.