Show that if $A\mathbf{x} = 0$... Let $$A = \begin{bmatrix}
1 & 1 & 1 \\
a & b & c \\
a^2 & b^2 & c^2
\end{bmatrix}.$$
Show that if $A\mathbf{x} = 0$ has a unique solution, then $a\neq b$, $b\neq c$ and $a\neq c$, i.e., $a,b,c$ are different.  
Well the determinant of this is $$\begin{align}\text{det}(A) &= -a^2b+ab^2+a^2c-b^2c-ac^2+bc^2 \\&= -(a-b)(a-c)(b-c).\end{align}$$ This means that when $a\neq b$, $b\neq c$ and $a\neq c$, then we have a non-zero determinant so $A\mathbf{x} = 0$ must therefore have a unique solution.
Is this correct?

Yes, your proof is correct. 
I hope you recognised that the matrix is Vandermonde, and that you didn't work out the determinant by hand!
