I am proving a topological statement by contradiction, and have made great progress. If I could just prove this next statement, I would be done:
For each line in the plane, select a non-trivial line segment. To each line segment, associate a infinite strip defined by the region between the lines perpendicular to the ends of the segment. Prove that there exists a disk of radius $1$ covered by the union of all strips no matter how we choose the line segments.
In the worst case scenario, all parallel lines produce the same segment, so we might as well see if we can find a disk using just lines passing through the origin. Unfortunately, I have no idea how to proceed after this.
I have a good news for you, the statement is true. Let’s prove it.
Let $\Bbb T$ be the unit circle centered at the origin. For each vector $v\in\Bbb T$ we can pick non-zero one-signed rational numbers $r’(v)<r’’(v)$ such that there exists an associated strip $S$ such that the borderlines of $S$ are perpendicular to $v$ and $S$ covers points $r’(v)v$ and $r’’(v)v$ (and so the whole segment between them). Baire theorem implies that there exist distinct points $v_1$ $v_2$ of $\Bbb T$, natural $n$, and rational numbers $r’<r’’$ such that a set $$D=\{v\in (v_1, v_2): r’(v)=r’,\, r’’(v)=r’’\}$$ is dense in $(v_1, v_2)$, where $(v_1, v_2)$ is an arc of an angle less than $\pi$ on $\Bbb T$ from $v_1$ to $v_2$. Now let $v$ be the midpoint of $(v_1, v_2)$, $r=\tfrac{r’+r’’}2$. Then each point $u$ which is outside the disk bounded by the circle $r\Bbb T$ and belongs to a cone $C$ emanating from a point $rv$ and bounded by the lines $rv-rv_1$ and $rv-rv_2$ (see the picture), is covered by an associated strip.
Indeed, it can be checked that there is a tangent from $u$ to an inner point $rw$ of an arc $(rv_1, rv_2)$. Thus $(u-rw,rw)=0$ so $(u,w)=r(w,w)=r$. By the continuity of the inner product, there exists $\varepsilon>0$ such that $|(u,w’)-r|<\tfrac{r’’-r’}2$ for each point $w’$ of $(v_1, v_2)$ such that the Euclidean distance $d(w’,w)$ from $w’$ to $w$ is less than $\varepsilon$. Since the set $D$ is dense in $(v_1, v_2)$, there exists a point $w’\in D$ such that $d(w’,w)<\varepsilon$. Since $r’(w’)=r’$, $r’’(w’)=r’’$, and $r’<(u,w’)<r’’$, we see that the point $u$ is covered by an associated strip.