Here is an independent set of 188 vertices, if I'm not mistaken. (At the end it is improved to 194...)

It seems more convenient to rename the vertices. Each vertex $(abc,de)$ gets a new name $[de,fgh]$ where $\{a,b,c,d,e,f,g,h\}=\{1,2,3,4,5,6,7,8\}$. Then two vertices $[de,fgh]$ and $[d'e',f'g'h']$ are connected if either $\{d,e\}=\{d',e'\}$ and $\{f,g,h\}\cap \{f',g',h'\}=\varnothing$, or $\{d,e\}\cap\{d',e'\}=\varnothing$ and $\{d,e,f,g,h\}=\{d',e',f',g',h'\}$. So, ten vertices with the same *underlying set* $\{d,e,f,g,h\}$ form a Petersen graph mentioned by Aaron Meyerowitz. We say that such 10-tuple of vertices is a *group*.

Now, take 46 of these groups such that their underlying sets $S$ contain 0, 1, or 2 elements from $\{6,7,8\}$. We choose from each such group four vertices; depending on $|S\cap\{6,7,8\}|$, these vertices have the following forms:

(0) $[1e,fgh]$ (so $\{e,f,g,h\}=\{2,3,4,5\}$);

(1) $[de,fgh]$, where $d\in\{6,7,8\}$ (so $\{e,f,g,h\}\subset\{1,\dots,5\}$);

(2) a bit more complicated: if the underlying set contains $\{6,7\}$, $\{7,8\}$, or $\{8,6\}$, then the vertices have respectively the forms $[6e,fgh]$, $[7e,fgh]$, or $[8e,fgh]$.

Now, it is clear that two chosen vertices with the same underlying set are not connected, since their pairs have a common element.

Assume that we have chosen two connected vertices whose underlying sets intersect by two elements. These may be either two vertices from (2), or a vertex from (2) and that from (1).

A vertex $[de,fgh]$ from (1) and a vertex $[d'e',f'g'h']$ from (2). Their underlying sets have two common elements from $\{1,2,3,4,5\}$, so their intersections with $\{6,7,8\}$ are disjoint. But each of the pairs $\{d,e\}$ and $\{d',e'\}$ contains an element from $\{6,7,8\}$, so these pairs are distinct.

Two vertices from (2): their underlying sets cannot contain the same pair of elements from $\{6,7,8\}$, otherwise these sets have at least three common elements. If, say, one set contains $\{6,7\}$ and the other contains $\{7,8\}$, then the first vertex has the form $[6e,fgh]$, while the second one cannot have such form.

**[EDIT]** It is more efficient to remove group (0) and to put instead 10 vertices of the form $[de,678]$ (no vertex $[d'e',f'g'h']$ from (1) and (2) has pair $\{d',e'\}\subset\{1,\dots,5\}$). This gives 194. Hope it's not optimal, otherwise I hardly imagine a proof of the optimality...

2more comments