Proving there's a set with the cardinality $\mathfrak c$ on the $x$ axis of points that do not belong to the set of disks 
Prove/disprove: On the $x$ axis there's a set with the cardinality $\mathfrak c$ of points that do not belong to any disk of a set $O$ of disjoint disks of positive radius $\{(x,y)\in \mathbb R|(x-a)^2+(y-b)^2\le r^2 \}:a,b,r\in \mathbb R: r \gt 0$.
Note: the disks can't overlap.

I think it's true, by sketching the disks I always get a gap between them, and even by filing these gaps to infinity they can't cover all the points on an interval because they can't overlap. It's impossible for any shape to not have gaps if they can't overlap.
In a previous part of the exercise I already found the cardinality of $O$ to be (at most) $\aleph_0$ by finding a rational point $(q_1,q_2)$ in each disk of $O: q_1,q_2\in \mathbb Q\times\mathbb Q$. So if we'll take all the numbers that do not belong to any disk, they would be a subset of the transcendental numbers and on the other hand, every disk would have two gaps from both sides of its perimeter, so we'll get $2^{\aleph_0}$.
 A: With the restriction that the closed disks do not overlap, we can prove the statement (Added, thanks Hagen) assuming the Continuum Hypothesis.  [NB: According to Asaf's Answer to this previous Question, we don't need to assume the Continuum Hypothesis for this particular case.  More added to the end of this Answer.]
Consider for the sake of contradiction that it was possible to cover all but at most countably many points of the $x$-axis with the closed disks.  Taking those points, together with the at most countably many nonempty intersections of the disks with the $x$-axis, we would have expressed the $x$-axis as a union of (at most) countably many disjoint bounded closed intervals (considering a single point $[s,s]$ as a closed interval here).  [NB: I put "at most" in parentheses before countably many because clearly any finite number of disjoint closed intervals would have compact union, and the $x$-axis is not compact.]
However it is not possible to do this.  According to a theorem of Sierpiński, if $\mathbb{R}$ or any Baire space is expressed as the countable union of disjoint closed sets, at most one of them is nonempty.
Since the whole $x$-axis cannot be realized as the intersection with one of the closed disks in $O$ or as a single point $[s,s]$, we are done.
Added:  Consider the intersection of those open sets which are complements of the closed disks' intersections in the $x$-axis (real line).  That intersection of countably many open sets is by definition a $G_\delta$ set.  By the argument above it is not countable, and by Asaf's Answer to Is the cardinality of uncountable $G_\delta$ set of $\mathbb{R}$ equals the cardinality of the continuum?, we conclude it has cardinality of the continuum without assuming the Continuum Hypothesis.
