Your solution for $X$ is correct.
Here is another solution (it is cheating, though, because I used Octave). On he plus side, it shows another approach to solving the equation.
$A$ is diagonalizable, and so has a basis of eigenvectors, $u_1,u_2$. Define the linear operator $L(X) = X-A X A^*$. You want to solve $L(X) = C$. Note that $u_i u_j^*$ forms a basis for the space of $2 \times 2 $ matrices, and $L(u_i u_j^*) = (1-\lambda_i \overline{\lambda_j}) u_i u_j^*$, hence $L$ is diagonalizable.
To express $C$ in terms of $u_i u_j^*$, let $C = \sum_{i,j} [\Gamma]_{ij} u_i u_j^*$, and solve for $\Gamma$. Let $U$ be the matrix whose columns are $u_1 ,u_2$. Note that $\sum_i [\Gamma]_{ij} u_i = U \Gamma e_j$, so we have $C = \sum_{j} U \Gamma e_j u_j^* = \sum_{j} U \Gamma e_j (U e_j)^* = \sum_{j} U \Gamma e_j e_j^T U^* = U \Gamma (\sum_{j} e_j e_j^T) U^* = U \Gamma U^*$. It follows that $\Gamma = U^{-1} C (U^{-1})^*$, and if we solve $L(\sum_{i,j} [\tilde{X}]_{ij} u_i u_j^*) = \sum_{i,j} [\Gamma]_{ij} u_i u_j^*$, we get $[\tilde{X}]_{ij} = \frac{1}{1-\lambda_i \overline{\lambda_j}} [\Gamma]_{ij}$.
Reversing the procedure to obtain $X$ in the standard basis, we have $X = U \tilde{X} U^*$.
Using Octave we obtain the same result as above.
a = [1.5 1 ; -0.7 0 ]
c = [1 0.5 ; 0.5 0.25 ]
[u,d] = eig(a)
l = [d(1,1) ; d(2,2) ]
phi = ones(2,2)-l*l'
gamma = inv(u)*c*inv(u')
x_tilde = (1./phi).*gamma
x = u*x_tilde*u'
# check result...
a*x*a'+c-x
Here is the output:
octave:44> a*x*a'+c-x
ans =
0.00000 - 0.00000i 0.00000 - 0.00000i
0.00000 - 0.00000i 0.00000 + 0.00000i
octave:45> x
x =
18.8802 + 0.0000i -11.3672 - 0.0000i
-11.3672 - 0.0000i 9.5013 + 0.0000i
I generally liked teaching out of an earlier edition of Poole (several years ago). I found I had time to do, and liked, some of his applications at the end of the sections. The students seemed to do well with the book. On the other hand (at least in the earlier editions--I haven't seen the more recent edition) the definition and discussion of general vector spaces was so late in the book that I had very little time for it. Also, the book was quite pricey at the time.
Best Answer
$A$ is clearly diagonalisable, as the eigenvalues of $A$ are $-1/2,1/2,1$. Therefore, $A = PDP^{-1}$, where
$$D = \begin{bmatrix} -0.5 & 0 & 0 \\[0.3em] 0 & 0.5 & 0 \\[0.3em] 0 & 0 & 1 \end{bmatrix}$$ Now, $A^n = PD^nP^{-1}$, therefore, $\displaystyle \lim_{n \rightarrow \infty} A^n = \lim_{n \rightarrow \infty} PD^nP^{-1} = P X P^{-1}$, where, $$X = \begin{bmatrix} 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 \\[0.3em] 0 & 0 & 1 \end{bmatrix}$$ Now, all that remains is to find $P,P^{-1}$. I'll leave this easy calculation, as the method is clear.