Part 1. Intuition. OK, it isn't very hard to just give a formal proof, but I'll try to explain this more intuitively first. If you have a set $M \subset \mathbb{R}$ with a supremum $m = \sup M$, then there are generally two possibilities: either $m \in M$, or $m \not \in M$. In the first case, $m$ is simply the maximum of $M$, and it isn't very interesting. On the other hand, in the second case $M$ doesn't have a maximum, but instead it has a lot of points "right below" $m$. For any positive $\varepsilon$ there should be a point $x$ in $M$ that is below $m$, but above $m-\varepsilon$. So the set $M$ looks very crowded in that area, with lots of points very close to each other.
And this is a clue to your problem. In your case, the set $M_x$ is very sparse, i.e. different points in $M_x$ are far apart from each other, because they are whole numbers. It means that $M_x$ isn't crowded at all, and the second case should be impossible. And in the first case we have $[x] = \sup M_x \in M_x$, which automatically means that $[x] \in \mathbb{Z}$.
Part2. A proof. Now we can build a proof using that intuition. Let $x \in \mathbb{R}$. Since $[x]=\sup M_x$, it follows that there exists an $y \in M_x$ such that $[x]-1 < y \leqslant [x]$. Note that $y$ is a whole number. Suppose that $y \neq [x]$, that is $y < [x]$. Then, by definition of the supremum, there exists an $y' \in M_x$ such that $y < y' \leqslant [x]$. And now we have $y,y' \in M_x$ such that $[x]-1 < y < y' \leqslant [x]$, which means that $|y'-y|<1$. But that is impossible, because $y$ and $y'$ are whole numbers.
This contradiction proves that $y = [x]$, and so $[x] \in M_x \subset \mathbb{Z}$, QED.
PS: yep, not the shortest way to do this. Then again, the definition of $[x]$ in the problem statement using supremums instead of maximums is also kind of strange. But I think that this solution goes well with that definition)