Consider the case when $OP$ makes an angle $\theta$ such that $0\leq\theta\leq\frac{\pi}{4}$. The argument then extends to all of $C$ by symmetry. Let $k$ be given. Consider all the lattice points of the form $Q(k,y)$ with $0\leq y\leq k$. Then by drawing all the rays $OQ$, we separate the eighth-circle into $k$ arcs. The length of the $j^{th}$ arc is $\tan^{-1}\left(\frac{j}{k}\right)-\tan^{-1}\left(\frac{j-1}{k}\right)$ by the formula $s=r\theta$, with $r=1$ in this case. Since $\tan x$ is concave down, i.e. $\frac{d^{2}}{dx^{2}}\tan x \leq 0$, when $x\geq 0$, the longest arc will be the first one, between the first and second rays, formed by the origin and $(k,0)$ and $(k,1)$, respectively. This arc has length $\tan^{-1}\left(\frac{1}{k}\right)-\tan^{-1}\left(\frac{1-1}{k}\right)=\tan^{-1}\left(\frac{1}{k}\right)$. Since $\tan^{-1}x<x$ for all $x>0$, $\tan^{-1}\left(\frac{1}{k}\right)<\frac{1}{k}$. So all the arcs have length less than $\frac{1}{k}$. This means that no matter what arc $P$ lies in, the length of the arc between $P$and the closest $Q'$ must be less than $\frac{1}{2k}$, which implies $PQ'<\frac{1}{2k}$, as required.