if $X,Y$ are two independent normal random variables, we know that $$X+Y$$ is still $N(u_x+u_y, \sigma^2_x + \sigma^2_y)$, and that's fine. (I proved it via characteristic function)
Now I want to find the distribution of $X-Y$.
By independence, and looking at the generating function:
$$E[e^{z(X-Y)}]= E[e^{zX} e^{zY}] = E[e^{zX}] E[e^{-zY}] = e^{z(u_x - u_y)} \cdot e^{\frac{z^2}{2} (\sigma_x^2 + \sigma_y^2)}$$
and since this is the moment generating function of a $N(u_x - u_y, \sigma_x^2 + \sigma_y^2)$ I conclude that $$X-Y$$ is distributed as $N(u_x - u_y, \sigma_x^2 + \sigma_y^2)$
Is my argument okay?
Your argument is fine.
You could also derive it from the fact you already know about the distribution of $X+Y$: just write $X-Y$ as $X+(-Y)$, and note that $X, -Y$ are independent normal random variables, where $-Y$ has mean $-u_y$ and variance $\sigma_y^2$