Suppose $\mathcal{P}$ is a partially ordered set. To $\mathcal{P}$ we can associate a simplicial complex $K(\mathcal{P})$ whose $n$-simplices are the chains of length $n+1$ in $\mathcal{P}$. Since subchains of chains are again chains, it follows that $K(\mathcal{P})$ is indeed a simplicial complex.
We can also define $K(\mathcal{P})$ to be the clique complex of the graph $G(\mathcal{P})$ whose vertices are the elements of $\mathcal{P}$, with edges between all pairs of vertices $x,y\in\mathcal{P}$ with $x<y$.
To each simplicial complex $K$, we have the associated topological space $|K|$, also referred to as the geometric realization of $K$.
Now if $\mathcal{Q}$ is a subposet of $\mathcal{P}$, then $|K(\mathcal{Q})|$ is a subspace of $|K(\mathcal{P})|$. In general, subspaces of contractible spaces need not be contractible, but in this specific case, does contractibility of $|K(\mathcal{P})|$ imply that of $|K(\mathcal{Q})|$?
I'm still trying to develop intuition for concepts such as contractibility and the geometric realization of a simplicial complex, so any answers that help me think about these concepts are much appreciated. Perhaps we can state contractibility in terms of the graph $G(\mathcal{P})$, and translate this question to a problem in graph theory.
No.
Observe that $|K(\mathcal{P})|$ is contractible whenever $\mathcal{P}$ has a minimum element, since you can contract everything to that vertex.
So let $\mathcal{Q}$ be any poset such that $|K(\mathcal{Q})|$ is not contractible, and let $\mathcal{P}$ be the poset obtained from $\mathcal{Q}$ by adding a minimum element. Then $|K(\mathcal{P})|$ is contractible, but $|K(\mathcal{Q})|$ is not.
For example, let $\mathcal{Q} = \{1,2,3,4\}$ with order $$ 1<3,\quad 1<4,\quad 2<3,\quad\text{and}\quad 2<4, $$ and let $\mathcal{P} = \{0,1,2,3,4\}$, where $0<q$ for all $q\in\mathcal{Q}$. Then $|K(\mathcal{Q})|$ is a square of edges, and $|K(\mathcal{P})|$ is the cone $|K(\mathcal{Q})|$, i.e. the four triangular faces of a square pyramid.