Sorry for two answers but in response to your comment:
"when I google these concepts it's too abstract, too vague,"
and
" I thought closed meant [a, b] but then we have [7, infinity) as closed."
I hope I can give an intuitive idea of "open" and "closed" and "bounded" and although in formal, tha is not misleading, and that in conjunction with a formal definition will render the formal definition less vague.
We have a universal space $X$. And for every point of $x$ we can draw little balls around this point. These balls can be as tiny as we want.
A set is a collection of points. Set can be connected or separate or any set of points.
A set $A$ is open: if for every point $a$ of $A$ we can draw a tiny ball around $a$ and all the points in the tiny ball will be in $A$.
So for example if $A = (0, 1)$ and we pick a point in $A$, say $a = 0.000001$. We can draw a tiny ball around $0.000001$ so that all points in the ball are in $A$. Because $a$ is so close to zero, the radius of this tiny ball must be very small. In fact it must be smaller than $0.0000005$. But we can do a ball and we can do a ball around every point. So $A$ is open.
As every point will have a ball around it, that means for every point there "a little further you can go". So $A$ will not have any sharp edges.
Now it is possible that there is a point $x$, (or more points) in the space $X$ that may, or may not be in $A$ and has the property, that every ball around $x$ will have some points in $A$.
For instance, take $A= (0,1)$ and $x = 0$. Notice every ball around $0$ will have some points that are bigger than $0$ (and smaller than $1$) and these will be in $A$. That will happen for every ball around $0$. It will also happen for every ball around $1$. And it will also happen for every ball around $0.75$ (or any point $y$ so that $0\le y \le 1$).
It will not happen for the point $-0.000001$ because we can take a tiny ball around $-0.0000001$ that completely misses $A$. But no ball around $0$ or $1$ can miss $A$.
We call such points "limit points" of $A$. Limit point: If every ball around $x$ must "hit" the set $A$ then $x$ is a limit point of $A$.
Notice $0.75$ is a limit point of $(0,1)$ and $0.75$ is in $(0,1)$ and notice $0,1$ are limit points of $(0,1)$ and $0$ and $1$ are not in $(0,1)$. If we were to list all the limit points of $A= (0,1)$ we would find they are the point $0$ and $1$ and all the points in between.
A set is closed: All its limit points are also points in the set.
So $(0,1)$ is not closed. Because the limit points $0$ and $1$ are not in the set.
But the set $B=[0,1]$ is closed. The only points that must hit $B$ when we draw balls around them are the points $[0,1]$, and those points are all in $B$.
Notice the space $X$ is both open and closed. For every point of $X$, every ball will be entirely in $X$ (there's nowhere else to be) so $X$ is open. Every ball around every point "hits" $X$ so every point is a limit point of $X$ and every point is in $X$ so $X$ is closed.
Notice not all sets have limit points. Take $\mathbb Z\subset \mathbb R$. If you take any real number you can draw a small ball around it and not hit any integer. So no point is a limit point of $\mathbb Z$.
This actually means that $\mathbb Z$ is closed. It doesn't have any limit points so all the limit points (all zero of them) are in the set so.... it is closed. It might be easier to say, there aren't any limit points that are not in $\mathbb Z$.
Notice the empty set is both closed and open. No points have balls that hit the empty set (there is nothing to hit) so there aren't any limit points of the empty set. So there aren't any limit points that are not in the empty set. So the empty set is closed.
The "empty set is closed" is a little more abstract. There aren't any points in the empty set that you can't draw a ball around and be entirely in the empty set. So for all the points that are in the empty st (all zero of them) you can't claim you *can't draw a ball around them them that is entirely in the empty set So the empty set is open.
Okay, bounded.... Well, bounded is exactly what it sounds like. Any distance between two points is finite. I'm not sure there is really anything more to say. The empty set is bounded because there is no distance between any two points.
Best Answer
No, the set described does not cover $M=[0,2]\cap\Bbb Q$. There are two ways to look at compactness of a subset $S$ of a topological space, which turn out to be equivalent (as you should prove to yourself): one is compactness of $S$ when considered under the subspace topology, and the other is the statement that any set of open sets in the space whose union includes $S$ as a subset (also called an open cover of $S$) has a finite subset whose union also includes $S$ as a subset. Thus you don't need to think about open sets that lie in $S$ but rather more broadly.
Consider the following open cover of $M$: $$\mathcal A=\{(-1,\sqrt 2)\}\cup\{(\sqrt 2+1/x,3)\mid x\in \Bbb R_+\}.$$