I haven't been able to find a solid proof for the standard Lyapunov theorem related to the stability of discrete systems. Every proof I've read seems to mimic the proof for the continuous case.
As an example, let's take a look at the proof given in this technical report, focusing on the underlined sentence (note that there is a mistake there, as it should have been $t \geq 0$ instead of $t \leq 0$ ).
I don't think this really proves that any trajectory starting in $\Omega_{\beta}$ stays in $\Omega_{\beta}$. Note that nothing prevents that, at some $t=k$, $||x(k)|| > r$, with $V(x(k)) \leq \beta$. For the continuous case the continuity of the trajectory $x(t)$ guarantees that if $||x(t_0)|| < r $ and $||x(t_1)|| > r $, then $||x(t_*)|| = r$ for some $t_* \in (t_1, t_2)$. From there we can see a contradiction, as $V(||x(t_*)||) \geq \alpha > \beta$. However, in the discrete case we can't have that certainty, as the "trajectory" is no longer continuous.
You are correct that you can't conclude that from the definition of $\Omega_{\beta}$, at least I don't see how. The main problem is that $\Omega_{\beta} = B_r \cap V^{-1}([0, \beta])$ could have several connected components.
You can fix this argument in the following way. Consider $C_{\beta} \subset \Omega_{\beta}$ to be the connected component of $\Omega_{\beta}$ containing $x=0$. Now we can prove that $f^n(C_{\beta}) \subset C_{\beta}$ for every $n\geq 0$.
The argument is very similar to the one you mentioned for the case of flows, but using the fact that $f^n(C_{\beta})$ is connected for every $n\geq 0$ and contains $x=0$, since it is a fixed point of $f$. First you prove that $f^n(C_{\beta}) \subset B_r$, for every $n\geq 0$. If it was not the case, then you can find a point $y\in C_{\beta}$ such that $\|f^N(y)\|=r$ for some $N>0$. Since the Lyapunov function is non-increasing along the orbit of $y$ this gives a contradiction with the choice of $\beta$.
Thus $f^n(C_{\beta})$ is connected, contains $x=0$ and is contained in $B_r\cap V^{-1}([0,\beta])$, which implies that $f^n(C_{\beta}) \subset C_{\beta}$. The rest of the proof follows in the same way as in your post.