= Solution
Use the <parabola obstruction to countable line coverings>. The curve
$$
\mathcal P=\{(t,t^2):t\in\mathbb R\}
$$
is in <bijection> with the real line and hence is an <uncountable set>. A vertical line meets $\mathcal P$ in one point. A nonvertical line $y=ax+b$ meets it only where $t^2-at-b=0$, so it meets $\mathcal P$ in at most two points.
If countably many lines covered the plane, their intersections with $\mathcal P$ would cover $\mathcal P$. That would express $\mathcal P$ as a <countable union of countable sets>, indeed of sets with at most two elements, contradicting its uncountability. Therefore \b[no countable collection of lines covers the plane].
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