Two points no matter how you choose from the six points in the unit disk are at distance at most 1? Six points are to be chosen in a unit disk ($x^2 +y^2 \leq 1$) , such that distance between any two points is greater than 1? I am unable to, I think I want to prove formally that no matter how the six points are chosen, there are two points with distance at most 1 from each other. 
For seven points, I think pigeonhole principle could be used to prove there is no such arrangement. We can easily choose seven points such that maximum distance between any two points is equal to 1. Something like this in the figure. Now you can't increase the distance between the points(orange) chosen on the circle as the side length itself is one.
 A: Apply the same argument as for 7 points (from the question statement and Ross Millikan comment), except make sure to draw the 6 sectors such that one of the 3 boundary diameters contains one of the 6 points. That is, draw the 6 sectors after you know the points, such that one of the points, say $P_1$, lies on one of the boundaries.
If any points are in either of the two neighboring sectors to $P_1$, they are at distance $\le 1$ from $P_1$. If not, by the pigeonhole principle, one of the 4 other sectors contains 2 of the 5 other points. And those two points are a distance of at most 1 from each other. So we're done.
Note: I am not sure how to cleanly prove the fact that two points in a $\pi/3$-sector are a distance of at most 1 apart. I have a way that I think works using law of cosines, but it seems peripheral to the question & a pretty inelegant argument, so not including it for now.

EDIT: Re:your comment OP, yeah I also don't have a nice way to make fully convincing/clean the intuition that no 2 points in a sector have distance $> 1$.
Here's the inelegant law of cosines way mentioned above which shows that two points in a sector have distance $\le 1$:
We're given two points $P_1$ and $P_2$ in one $\pi/3$-sector. Say the two points are at a distance $r_1$ and $r_2$, respectively, from the center. Let the central angle between them be $\theta \le \pi/3$.
By the law of cosines, the square of the distance between $P_1$ and $P_2$ is
$$d(P_1, P_2)^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta \le r_1^2 + r_2^2 - 2r_1r_2(\frac{1}{2}) = r_1^2 + r_2^2 - r_1r_2$$
because $r_1, r_2 \ge 0$ and $0 \le \theta \le \pi/3$.
Now we have to show $r_1^2 + r_2^2 - r_1r_2 \le 1$ for $0 \le r_1, r_2, \le 1$.
There's probably be a nicer way to proceed here too, but here's one argument: we have $r_1^2 + r_2^2 - r_1r_2 = (r_1 - \frac{1}{2}r_2)^2 + \frac{3}{4}r_2^2$.
With $0 \le r_1, r_2, \le 1$, for fixed $r_2$, this expression is maximized when $r_1 = 1$ (since $r_1=1$ maximizes $|r_1 - \frac{1}{2}r_2|$ for every $0 \le r_2 \le 1$, uniquely except at $r_2=1$ when $r_1=0$ is also a maximizer).
So, for $0 \le r_1, r_2, \le 1$,
$$r_1^2 + r_2^2 - r_1r_2 = \left(r_1 - \frac{1}{2}r_2\right)^2 + \frac{3}{4}r_2^2 \le \left(1 - \frac{1}{2}r_2\right)^2 + \frac{3}{4}r_2^2 \\= 1 - r_2 + r_2^2 = \left(r_2 - \frac{1}{2}\right)^2 + \frac{3}{4} \le \frac{1}{4} + \frac{3}{4} = 1$$
(maximum achieved at $r_2 = 0$ or $r_2=1$ when $r_1 = 1$). 
