Showing a matrix is not diagonalizable

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.

The eigenvalues are correct, they are $\lambda_1=1$ and $\lambda_2=4$. The algebraic multiplicities are $m(\lambda_1)=2$ and $m(\lambda_2)=1$.

The eigenspace relative to $\lambda_1$ is (as you have correctly found) $V_{\lambda_1} = \langle (1,0,0) \rangle$. The eigenspace relative to $\lambda_2$ is $V_{\lambda_2} = \langle (1,3,9) \rangle$.

As you said, since the $1=\dim V_{\lambda_1} \ne m(\lambda_1) =2$ you can easily conclude that the matrix is not diagonalizable.