here's my try:

Let $X$ be a countable set of lines in the plane. the cardinality of the set of all lines in the plane with a slope between $0$ and $2\pi$ is $\aleph$ so there must be some line in the plane, $\Gamma$, with a slope $\alpha$ that is not in $X$. so now, every line in $X$ must intersect $\Gamma$ at only one point. because $X$ is a countable set then there is only a countable number of points on $\Gamma$ that the lines of $X$ cover so there is an uncountable set of points on $\Gamma$ that the lines of $X$ don't cover (because $|\Gamma |=\aleph$). Hence, the union of lines in $X$ don't cover the plane.

first of all, is this proof correct? and second, can anyone give me another proof, maybe an easier one?