Showing a matrix is not diagonalizable 
For eigenvalue 1, the eigenspace I got was $span[ 1,0,0]$, and for eigenvalue 4, the eigenspace I got was $span[ 1,-3,9]$. Do they look right?
The reason the matrix is not diagonalizable is because we only have 2 linearly independent eigevectors so we can't span R3 with them, hence we can't create a matrix E with the eigenvectors as its basis. Is that correct, did I word it right?
 A: 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.
A: 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.
