CTFT to DTFT why can't you always just substitute $\Omega = \omega/T_S$
It is due to the property of unit impulse function

\begin{equation}
\delta(\alpha t)  = \frac{ 1}{|\alpha|} \delta(t) 
\end{equation}

(Usually $\omega $ is used for angular frequency and $\Omega $ is used for angle. Interpret my answer accordingly. )

suppose 
\begin{equation}
X_a(\omega) = \delta(\omega -x_0 )  
\end{equation}
and if we substitute '$\omega$' with $\frac{\Omega}{T_s}$ then 
\begin{equation}
X_a(\frac{\Omega}{T_s}) = \delta(\frac{\Omega}{T_s}  -x_0 )
   = \delta(\frac{\Omega - x_0 T_s}{T_s} )
   =  \delta(\frac{ 1}{T_s} (\Omega - x_0 T_s) )
   =  {T_s} *\delta(\Omega - x_0 T_s) 
\end{equation}
where '$\alpha$'  is $\frac{ 1}{T_s}$ 
and 
$$
\frac{ 1}{|\alpha|} = T_s 
$$

and when you do DTFT calculation on 

\begin{equation}
X(\Omega)  = \frac{ 1}{T_s} X_a(\frac{\Omega}{T_s})
   =\frac{ 1}{T_s} * ( T_s * \delta(\Omega - x_0 T_s) )
   = \delta(\Omega - x_0 T_s) 
\end{equation}

in the final answer $T_s$ and $\frac{ 1}{T_s}$ cancels 

and you will get the expression based on the independent variable '$\Omega - x_0 T_s$'

So there is nothing wrong in your actual answer.