Set of diameter $\le 1$ contained in set of constant width $1$ I'm reading the paper Minimal universal covers in $E^n$ by H.G. Eggleston and they state that every set $A\subseteq{\bf R}^2$ of diameter at most $1$ (the diameter of $A$ is defined as $\sup_{x,y\in A}|x-y|$) is contained in a set which has width $1$ along any direction.  I can't see how the proof of this should work, though.
EDIT: Maybe the proof is something along the lines of constructing a thing that has width $1$ along any rational direction, then it should has width $1$ along all directions.  For this, it should suffice to show that for any $A$ of diameter $\le 1$ you can add a translate $I$ of the unit interval (embedded into ${\bf R}^2$) such that $A\cup I$ has diameter $1$.  Then you are able to add an infinite amount of intervals at different angles without the diameter exceeding $1$.
 A: Let $X\subseteq\mathbf{R}^2$ be of diameter $d(X)\le 1$, w.l.o.g. we can assume $X$ to be closed.  I attempt to show that there exists $a\in\mathbf{R}^2$ such that with $b:=a+(1,0)$ we have $d(X\subseteq\{a,b\})\le 1$.  This way we can add pairs of points with distance $1$ along every rational angle, which should suffice to get a superset of $X$ with constant width $1$.  In the following, $b$, $b'$, etc. always means $a+(1,0)$, $a'+(1,0)$, etc.
[EDIT] Look in the comments for a shorter approach that doesn't use the following section.
First choose $a$ such that we have $$d(X\cup\{a,b\})=\min_{a'\in\mathbf{R}}d(X\cup\{a',b'\}),$$ call this diameter $D$.  Assume there is some $a'\neq a$ with the same property.  Then all points on the line between $a$ and $a'$ must have this property as well, as when you travel in a direction and the distance to some point grows, it will not shrink again when you travel further in the same direction.  Assume $1<D$.  Then there exists $x\in X$ such that $D$ equals the distance from $x$ to $(a+a')/2$ (or just any other point strictly between $a$ and $a'$).  But this implies $D<d(x,a)$ or $D<d(x,a')$, a contradiction.  So if $1<D$ then our $a$ must be unique.
Now assume w.l.o.g. that $a=(-1/2,0)$, $b=(1/2,0)$.  Consider the set $E$ of points $e$ such that $D=d(\{a,b,e\})$, I called it $E$ because it has the shape of a vertical eye, note that $X$ must live in $E$ together with its enclosed area.  For $s,s'\in\{-,+,*\}$ denote by $E_{s,s'}$ the set of points $e\in E$ such that the sign of the first component is $s$ or $0$ or anything if $s=*$ and similar for the sign of the second component with $s'$.  When we shift $a$ to the right a little, the diameter of $X\cup\{a,b\}$ increases, so $X\cap(E_{-,*})$ cannot be empty.  A similar argument can be made for $E_{+,*}$ and by shifting $a$ vertically also for $E_{*,-}$ and $E_{*,+}$.  So there must be two points of $X$ in two opposite components of $E$ (i.e. $E_{-,+}$ / $E_{+,-}$ or $E_{-,-}$ / $E_{+,+}$), and one can easily see that those have distance greater than $1$, a contradiction.
A: WLOG we can assume that $X$ is convex. $p_1,\
p_2\in X$ s.t. $|p_1-p_2|=1$. Then there is a circular sector $V_1$ at $p_1$ whose angle is $\theta_1$.
And $V_1$ contains arc $p_2p_3$. Here $\theta_1$ is the largest when $V_1-X$ is empty.
Now we do the same thing at $p_2$. Hence we have $\sum_{i=1}^{2m+1}
\ \theta_i =\pi,\ \theta_i>0$ and $\bigcup_{i=1}^{2m+1} \ V_i$ contains $X$.
