The algebraic multiplicity of $\lambda=1$ is $2$. A matrix is diagonalizable if and only if the algebraic multiplicity equals the geometric multiplicity of each eigenvalues.
By your computations, the eigenspace of $\lambda=1$ has dimension $1$; that is, the geometric multiplicity of $\lambda=1$ is $1$, and so strictly smaller than its algebraic multiplicity. Therefore, $A$ is not diagonalizable.
Note that you don't actually need to compute the eigenspace to determine diagonalizability: you just need to figure out the dimension of the eigenspace. The eigenspace of $\lambda=1$ is the nullspace of $A-I$. Since $A-I = \left(\begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 3 \end{array}\right)$ has rank $2$, it has nullity $1$, so the dimension of the eigenspace corresponding to $\lambda=1$ is $1$, strictly smaller than the algebraic multiplicity. This suffices to show $A$ is not diagonalizable.