This text is an extended *Answer* that I have posted some 17h ago and removed some 16h ago.

**THE NEW PART**: I've removed my answer because @PietroMajer has posted earlier (some 19h ago) an elegant 9-point example that had a richer and more impressive structure (and Pietro has posted a comment under the OP too). This 9-point solution contained a 6-point solution.

However, I've decided to mention the existence of that 6-point solution explicitly. Of the Pietro's 9 points one may take only 6-points

$$ \{\xi^k+c\cdot\xi^j: k=0,1,2\ \ \text{and}\ \ j=0,1\} $$

**THE OLD PART** (posted about 3h **after** @PietroMajer's solution):

In $\ \mathbb C,\ $ let $a$ and $b$ and $c$ be the vertices of a triangle such that $\ |a-c|=|b-a|=|c-a|=1.\ $ Let $\ v\in\mathbb C\ $ be such that
$\ |v|=1,\ $ and $\ \pm v\ $ be different from any $\ a-c\ $ or $\ b-a\ $ or
$\ c-a.\ $ Then the $\ 6$-point set consisting of points $\ x+v,\ $ where $\ x\ $ is any of the points $\ a\ $ or $\ b\ $ or $\ c\ $ together with these three points, provides an example of a $\ 6$-point planar set such that each point has distance $\ 1\ $ from at least three other selected points.

Is $\ 6\ $ the record or is it $\ 5?$ — I feel lazy :) but this should not be a difficult question to settle.

Oh, $\ 5\ $ is impossible. This is a simple exercise. Thus, the record is $\ \mathbf 6$.

**PS**. 5-point solution is impossible even if collinearity was allowed.

generic). $\endgroup$9more comments