countable family of open sets

Yes. We can take the countable many open intervals with rational center $\frac ab$ and radius $\frac1{b^2}$, $$U_{a,b}:=\left]\frac ab-\frac1{b^2},\frac ab-\frac1{b^2}\right[,$$ with $a\in\Bbb N_0$, $b\in \Bbb N$, $a\le b$, $\gcd(a,b)=1$.

If $\frac ab\in[0,1]$ is rational, then the distance to any other rational $\frac cd$ is at least $\frac1{bd}$ and this is $>\frac1{d^2}$ for almost all $\frac cd$, hence $\frac ab\in U_{c,d}$ at most for the finitely many cases when $0\le c\le d\le b$.

On the other hand, if $\alpha\in[0,1]$ is irraional, then we find infinitely many $a,b,c,d$ with $$\tag1\frac ab<\alpha<\frac cd\quad\text{and}\quad ad-bc=-1, \text{ i.e., }\frac cd-\frac ab=\frac1{bd}-$$ More concretely, we can take $(a_0,b_0,c_0,d_0)=(0,1,1,1)$ and if we have $(a_n,b_n,c_n,d_n)$ such that $(1)$ holds, then either $(a_{n+1},b_{n+1},c_{n+1},d_{n+1}):=(a_n,b_n,a_n+c_n,b_n+d_n)$ or $(a_{n+1},b_{n+1},c_{n+1},d_{n+1}):=(a_n+c_n,b_n+d_n,c_n,d_n)$ works as next choice, depending on whether $\alpha<\frac{a_n+c_n}{b_n+d_n}$ or $\alpha>\frac{a_n+c_n}{b_n+d_n}$ (equality cannot occur) As $$\frac cd-\frac ab=\frac 1{bd}<\frac1{b^2}+\frac 1{d^2},$$ we conclude that $$\tag2\alpha\in U_{a_n,b_n}\cup U_{c_n,d_n}$$ for each of our tuples. And as $\frac1{b_nd_n}\to 0$, we have both $\frac{a_n}{b_n}\to \alpha$ and $\frac{c_n}{d_n}\to \alpha$, hence no $U_{a_n,b_n}$ or $U_{c_n,d_n}$ occurs infinityly often in $(2)$.

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Alternative use of the same idea, but perhaps with nicer proof:

For every quadruple $(a,b,c,d)\in\Bbb N_0^4$ with $b,d\ge1$ and $ad-bc=-1$ let $$U_{a,b,c,d} =\left]\frac ab,\frac cd\right[.$$ Assume $\frac xy$ is a rational number in $U_{a,b,c,d}$. Then from $\frac ab<\frac xy$, we have $\frac xy-\frac ab=\frac{bx-ay}{by}>0$, hence for the numerator $bx-ay\ge 1 $ and simimlarly we find $yc-xd\ge1. $ Hence $$y=(bc-ad)y=d(bx-ay)+b(yc-xd)\ge b+d.$$ It follows that a rational number $\frac xy\in[0,1]$ can only be in the finitely many $U_{a,b,c,d}$ with $b+d\le y$, $0\le a<b$, $1\le c\le d$.

On the other hand, each irrational $\alpha\in[0,1]$ is certainly $\in U_{0,1,1,1}$. If there were only finitely many of such open sets containing $\alpha$, then there'd be one with maximal $b+d$. But one verifies that $$U_{a,b,c,d}=U_{a,b,a+c,b+d}\cup U_{a+c,b+d,c,d}\cup\left\{\frac {a+c}{b+d}\right\} $$ and hence $\alpha$ must be in one of the intervals on the right, contradicting maximality.

It's a fact (which I'm not going to use except as motivation) that any two countable dense subsets of $\mathbb R$ are "equivalent" in the sense that there is a homeomorphism of $\mathbb R$ with itself that maps one to the other. Therefore, if the statement you want to prove is true, it remains true if the set $\mathbb Q$ of all rational numbers is replaced with any other countable dense subset of $\mathbb R$. Hence it seems natural to seek a proof which is set-theoretical rather than arithmetical.

In fact, the set doesn't even have to be dense; if $A$ is any countable subset of $\mathbb R$, there is a countable family $\mathcal U$ of open sets such that no element of $A$, but every element of $\mathbb R\setminus A$, belongs to infinitely many members of $\mathcal U$.

Case 1. $A$ is finite.

Let $U$ be the collection of all rational open intervals which are disjoint from $A$.

Case 2. $A$ is countably infinite.

Let $A=\{a_n:n\in\mathbb N\}$ (enumerated without repetition) and let $\mathcal U=\{U_n:n\in\mathbb N\}$ where $U_n=\mathbb R\setminus\{a_1,\dots,a_n\}$.

The statement is still true, but slightly less trivial, if you want the members of $\mathcal U$ to be bounded open intervals. Suppose $A=\{a_n:n\in\mathbb N\}$. Let $$\mathcal U=\mathcal W\cup\bigcup_{n\in\mathbb N}\mathcal V_n$$ where $\mathcal W$ is the set of all rational open intervals disjoint from $A$, and $\mathcal V_n$ is the set consisting of the bounded components of $\mathbb R\setminus\{a_1,\dots,a_n\}$ together with the intervals $(L_n-1,L_n)$ and $(R_n,R_n+1)$ where $L_n=\min\{a_1,\dots,a_n\}$ and $R_n=\max\{a_1,\dots,a_n\}$. It is easy to see that $\{U\in\mathcal U:x\in U\}$ is finite if $x\in A$, infinite if $x\notin A$.