TITLE: How well path-connected can a (special) partition of $\Bbb R^2$ be?
QUESTION [5 upvotes]: It turned out that How well connected can a (special) partition of $\Bbb R^2$ be? had a a few nice answers using continum-many pairwise disjoint dense connected sets. Connectedness is just weirder than one first thinks ...
So let's up the ante to path-connectedness:
Let $\{A_i\}_{i\in I}$ be a family of subsets of $\Bbb R^2$ (where $I=\Bbb N$ or $\Bbb Z$; I don't know if it makes a difference - in the linked question the distinction was irrelevant as  solutions with all indices "completely linked" were found even for $I\approx \Bbb R$) such that

$\bigcup_{i\in I} A_i=\Bbb R^2$
$i\ne j\implies A_i\cap A_j=\emptyset$
$A_i\ne\emptyset$
$A_i$ is path-connected
$A_i\cup A_{i+1}$ is path-connected

How often can it happen that $A_i\cup A_j$ is path-connected for $j\notin\{i-1,i,i+1\}$?
Definition. Let's say that an index  $n\in I$ is infinitely linked/almost completely linked/completely linked if $A_n\cup A_i$ is path-connected for infinitely many/almost all/all $i\in I$. 
One possibility configuration is to let the $A_i$ be vertical stripes, in which case no $n\in I$ is infinitely linked.
With another configuration one can achieve that  there exists exactly one $n\in I$ that is completely linked (let $A_n=\{(0,0)\}$ and all other $A_i$ suitable sectors of $\Bbb R^2\setminus\{(0,0)\}$).
Can more than that be achieved? 

I.e., are there configurations with more (two, three, arbitrarily many, infinitely many, almost all, all) completely linked indices? 
Or maybe at least more almost completely linked indices?
For $I=\Bbb Z$, I can find configurations two infinitely linked indices. Are three or more infinitely linked indices possible? 
Are two infinitely linked indices possible with $I=\Bbb N$?

Note that  with $\Bbb R^3$ instead of $\Bbb R^2$ one can easily achieve that all indices are completely linked.
EDIT: Only after seeing Crostul's solution (and how it can be adapted for $I=\Bbb Z$, I noticed that it is not possible to have three or more completely linked indices: Pick a point in each of the three sets and a point in each of three other sets. Then we can pick a path witnessing the path-connectedness of the respective unions (wlog.(!) these paths do not intersect) and obtain a solution to the famous gas-water-electricity problem (i.e., a planar embedding of $K_{3,3}$), which is impossible.

REPLY [4 votes]: This picture represents a partition $\{ A_n \}_{n \ge 1}$ with $A_1$ and $A_2$ completely linked.