Lets say that a set of lines of the real plane covers the plane if, for every element $\langle x,y\rangle\in\mathbb{R}^2$, there exists a line $l$ of the set that passes through $\langle x,y\rangle$.
Its obvious that the set of all lines of the plane, denoted by $\mathscr{L}$, has this property, and it can be identified with the real projective plane $\mathbb{P}_{\mathbb{R}}^2$ minus a point, say $(1:0:0)$. It is clear then that the cardinality of the set $\mathscr{L}$ is the same as the cardinality of $\mathbb{P}_{\mathbb{R}}^2$, but, how can we prove rigorously that every subset of $\mathscr{L}$ that covers the plane is actually equipotent with $\mathscr{L}$?
We know, for instance, that any line in $\mathbb{P}_{\mathbb{R}}^2$that doesn't pass through the distinguished point $(1:0:0)$, is a pincel of lines on the plane, and therefore, any line of $\mathbb{P}_{\mathbb{R}}^2$ that does not pass through $(1:0:0)$ can be identified with a set of lines that cover the plane.
I think this observation must be useful for the proof, but I don't know how to exploit this fact.
The thing is that any line of $\mathbb{P}_{\mathbb{R}}^2$ has the cardinality of $\mathbb{P}_{\mathbb{R}}^2$, and a pencil is one of the simplest subsets of $\mathscr{L}$ that cover the plane, because for every point $\langle x,y\rangle$ of the plane, there is only one line that passes through $\langle x,y\rangle$. "Intuitively", any subset of $\mathscr{L}$ that covers the plane must have a cardinal larger or equal to that of the pencil, because for a point $\langle x,y\rangle$ there might be more than one line that passes through it.
Since any pencil of lines through a point and the set $\mathscr{L}$ have the cardinality of $\mathbb{P}_{\mathbb{R}}^2$, from the previous commentary we should conclude that any two subsets of lines that cover the plane should have the same cardinality.
Hope someone helps me make a rigorous proof out of this intuitive and weak argumentation.
Thanks in advance for your time.
EDIT: just posted a full, complete answer to this question on the comments, after thinking a lot about this question. I hope that it is correct, and if not, I would really like to hear about your suggestions and comments.
Best Answer
The set of all lines in the plane has the cardinality of the continuum.
If you have a family of lines covering the plane but which omits a line $L$, then there must be for each $P\in L$ a line in the family containing $P$. These lines must all be distinct, so there are continuum many of them.