Here is a problem I've been trying to solve for some time now. Maybe you could help me. We have two sets
$\mathcal {S}$ is a family of circles in the plane such that for any $x \in \mathbb{R}$ there exists a circle $O \in \mathcal {S}$ which intersects $x$ axis in $(x,0)$.
$\mathcal {T}$ is a family of circles in the plane such that for any $x \in \mathbb{R}$ there exists a circle $O \in \mathcal {T}$ which is tangent to $x$ axis in $(x,0)$.
We need to show that in each of these sets there exist at least two different circles whose intersection isn't empty. What it comes down to is that we have to prove that $card (S) < card (\mathbb{R})$. Maybe we could somehow identify each circle with a different rational number?
Assuming what you want to prove is "Any set of circles, the union of which contains the entire $x$ axis contains two intersecting circles", then the truth of this statement depends on whether or not you admit degenerated circles of radius $0$ (isolated points) in the family. If you do, then this statement is false, trivialy because you can cover the $x$ axis by disjoint circles of radius $0$, or a bit more subtly because you can take the family of all circles with the origin as center. If you leave out the degenerated circle from the latter example, then you cover all of the $x$ axis except the origin itself. I think this example suggests that, even if you forbid degenerated circles, a bit more than a cardinality argument is required.
Here's a (corrected) proof assuming circles must have positive radius. First a simple
Lemma A family of disjoint circles, all tangent to the $x$ axis, is at most countable.
This follows because two disjoint circles tangent to the same line are boundaries of disjoint disks, each disk contain a point with rational coordinates, and there are only countably many such points.
This immediately establishes the statement about $\mathcal T$.
Now suppose you have a family $\mathcal S$ of disjoint circles whose union covers the $x$ axis. Pick a circle in $\mathcal S$ to start with that intersects the $x$-axis twice, which must exist due to the lemma. The interval between the point at $\frac13$ and $\frac23$ between the two points of intersection contains at most countably many points of tangency of a circle in $\mathcal S$ with the $x$ axis; therefore we can choose the circle in $\mathcal S$ passing through another point of this segment, and this cicle must intersect the $x$-axis twice. The two circles being disjoint, the second circle must bound a disk entirely contained in the disk bounded by the first circle, and its radius is less than $\frac23$ of the radius of the first circle. Now continue as before with the second circle, and so on forever. You get infinitely many circles in $\mathcal S$, of radius tending to $0$, where the disks that are bounded by them are each contained in the previous disk. The intersection of all these disks then is a singleton point $P$ that must lie on the $x$ axis (easy proof, its distance from the $x$ axis cannot be positive). Since $P$ is surrounded by arbitrarily small circles from $\mathcal S$, there does not exist any circle at all that passes through $P$ while remaining disjoint from the all those circles. Yet $\mathcal S$ must contain such a circle, giving a contradiction.