With

In[1]:= Clear[data]
data = {{{1}, {2}, {1, 2, 3}}, {{1}, {3}, {1, 2, 3}}, {{1}, {1, 
    2}, {1, 2, 3}}, {{1}, {2, 3}, {1, 2, 3}}, {{2}, {3}, {1, 2, 
    3}}, {{2}, {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1, 2, 3}}, {{3}, {1,
     2}, {1, 2, 3}}, {{3}, {2, 3}, {1, 2, 3}}, {{1, 2}, {2, 3}, {1, 2,
     3}}};

one implements the conditions straightforward

(* it is actually a conditional condition *)
Clear[wierzC1, wierzC]
wierzC1[l_List] := True /; Length[l] == 1
wierzC1[l_List] := (Length[l] - 
     Length[Select[
       Partition[l, 2, 
        1], ((#[[1, 1]] == #[[2, 1]]) || (#[[1, -1]] == #[[
             2, -1]])) &]] == 1) /; Length[l] > 1
(* preconditions:
p[i]: list of lists without sublists 
p[ii]: no empty sublists
p[iii]: length of the sublists in list is monoton
 *)
wierzC[l_List] := Block[{l0 = Length /@ l, sub1 = {}},
   If[Length[Cases[l0, 1]] > 1, (* condition 2 *)
    sub1 = Select[l, (Length[#] == 1) &];
    If[ContainsAny[
      Abs[Dot[{-1, 1}, #]] & /@ 
       First[Flatten[Tuples[sub1, {2}], {3}]], {1}],
     False, (* else: condition 1 for the rest *)
     wierzC1[Complement[l, sub1]]
     ], (* else: condition 1 *)
    wierzC1[l]
    ]
   ] /; (And @@ (VectorQ /@ 
       l)) && (And @@ (NonNegative /@ (Dot[{-1, 1}, #] & /@ 
         Partition[Length /@ l, 2, 1]))) && !ContainsAny[l, {{}}]

to reach the result

In[8]:= Select[data, wierzC]
Out[8]= {{{1}, {3}, {1, 2, 3}}, {{1}, {1, 2}, {1, 2, 3}}, {{2}, 
          {1, 2}, {1, 2, 3}}, {{2}, {2, 3}, {1, 2, 3}}, {{3}, {2, 3}, {1, 2, 3}}}

even in a language with an overwhelming rich set of keywords it is more often than not the case that a single word is not enough to let it do what you want it to do.

Crossposted here.

Thank you for your comment. Yes, you're correct. But then {{1},{3},{1,2,3}} would also be eliminated, by Rule 1, since 3 is not the first or last element of {1} and so this the example for why we need the Rule 2. Rule 1 has to hold for all consecutive elements in the the list.