Does the union of the subcomplexes of $X$ that contain a given subcomplex and whose inclusion in $X$ is trivial on $\pi_1$, have trivial $\pi_1$? Let $X$ be a simplicial complex and let $A \subset X$ be a contractible subcomplex on the same set of vertices as $X$. Is it true that the union $$\bigcup C$$ taken over all complexes $A \subset C \subset X$ whose inclusion in $X$ induces the trivial map on fundamental groups, has trivial fundamental group?
 A: I think it does not. Begin with $A$ a path $\{1,2\},\{2,3\},\{3,4\},\{4,5\}$. To construct $X$, add an edge $\{1,3\}$ and triangles $\{1,2,5\},\{1,3,5\},\{2,3,5\}$ to $A$.
$A$ is contractible. If $C$ contains $A$ and its image in $X$ has trivial fundamental group, it does not contain any of the edges $\{1,5\},\{2,5\},\{3,5\}$: for instance, if it contains $\{3,5\}$, then it contains the closed loop $3\to 4 \to 5 \to 3$ (a triangle) but there is no triangle containing the vertex $4$, so there is no nullhomotopy.
The path $1\to 2\to 3\to 1$ is contractible in $X$: the union of triangles we added is a disk which bounds it. Thus $\bigcup C$ of the question is just $A \cup \{1,3\}$. It contains none of the triangles of the complex: they each contain one of the forbidden edges $\{i,5\}$ where $i\in \{1,2,3\}$. So the path $1\to 2\to 3$ is not contractible in $\bigcup C$.
A: Take $X$ to be a circle with three 0-simplices and three 1-simplices. Fix a point, then the union of all contractible subcomplexes containing that point is the whole circle.
EDIT : taking into account the fact that $A$ needs to contain all vertices.
I think it is true. At some point the proof isn't totally formal but I think it can be made rigorous. Let $Y$ be the union of the subcomplexes $C$ with a trivial inclusion in fundamental group and containing $A$, and take a map $f:S^1\to Y$. By some strong version of simplicial approximation you can suppose that there is a simplicial structure on $S^1$ such that $f$ is homotopic to a simplicial inclusion. Take your chosen basepoint of $S^1$,  $*$, and run over the circle clockwise. The images of $f$ begin in some subcomplex $C$ of the union and you eventually quit it. Consider first point of the circle after which you quit $C$, say $p$. Then $f(*)$ and $f(p)$ can be linked by a path in $A$ because it contains all the points and is contractible. This path is also in $C$. Glue this path and the path drawn by $f$ from $*$ to $p$, then it makes a loop on $C$ which is trivial in the fundamental group of the whole space $X$. So, $f$ is homotopic in $X$ to the same loop but where you have replace the path from $f(*)$ to $f(p)$ by your chosen path in $A$. But we rather want a homotopy in $Y$. Nevermind : the homotopy can be simplicial approximated in the 2-skeleton of $X$ and you can add to $C$ the 2-cells needed for the homotopy. This part is less formal but I think it can be written well. Do that by induction until you run out of vertices of $S^1$, and then the new $f$ has values in $A$ so is nulhomotopic.
