Real Analysis Topology Problem

This kinda reminds of the proof that $\mathbb{R}$ is uncountable. Suppose $\mathcal{F}$ is countable. Then the set of sets from $\mathcal{F}$ which contain $\mathbb{Z}$ as a subset (let's call it $\mathcal{G}$) is also countable. Let's write $\mathcal{G}=\{F_n\}_{n=1}^\infty$. Now we are going to define a new sequence of sets.

First of all we define $A_0=(-\infty, \frac{1}{2})$.

We know that $1\in F_1$. Since $F_1$ is open there is $0<\epsilon<\frac{1}{2}$ such that $(1-\epsilon,1+\epsilon)\subseteq F_1$. So now define $A_1=(1-\frac{\epsilon}{2},1+\frac{\epsilon}{2})$.

Next, as we know $2\in F_2$. Since it is an open set there is $0<\epsilon<\frac{1}{2}$ such that $(2-\epsilon,2+\epsilon)\subseteq F_2$. So now define $A_2=(2-\frac{\epsilon}{2},2+\frac{\epsilon}{2})$.

Continue this way by induction and finally define $A=\cup_{n=0}^\infty A_n$. Then $A$ is an open set as a union of open sets, and it contains $\mathbb{Z}$. Hence it must contain a set from $\mathcal{G}$. However, does it contain $F_1$? The answer is no, because $1+\frac{2\epsilon}{3}$ is an element in $F_1$, but not in $A$. And in a similar way you can show that $A$ doesn't contain any of the sets $F_n$. This is a contradiction.