As you say, a poset with a top or bottom element is contractible.
But more is true:
Lemma 1.
A preordered set such that every finite subset admits an upper or lower bound is weakly contractible, i.e. the geometric realisation of its nerve is contractible.
Thus, although $Q$ does not have a top or bottom element, the fact that every finite subset has an upper bound still makes $Q$ weakly contractible.
(Note that any upper bound will suffice – it does not even need to be a least upper bound!)
The above lemma is a special case of the following:
Lemma 2.
A category such that every finite diagram admits a cone or cocone is weakly contractible.
Even more generally:
Lemma 3.
A simplicial set $X$ with the following property is weakly contractible:
- For every finite simplicial set $K$ and every morphism $x : K \to X$, there exist a contractible simplicial set $L$, a monomorphism $j : K \to L$, and a morphism $\bar{x} : L \to X$ such that $\bar{x} \circ j = x$.
A completely rigorous proof is a bit messy but the idea is straightforward: we just have to show that for every $n \ge 0$ and every continuous map $x : S^n \to \lvert X \rvert$, there is a continuous map $\bar{x} : D^{n+1} \to \lvert X \rvert$ extending $x$ along the inclusion $S^n \hookrightarrow D^{n+1}$.
This implies the homotopy groups of $X$ are all trivial, so (by Whitehead's theorem) $X$ is contractible.
The messy part is to show that there is no loss of generality in assuming that there is a finite simplicial set $K$ and a morphism $K \to X$ such that we have a homeomorphism $\lvert K \rvert \cong S^n$ identifying $\lvert K \rvert \to \lvert X \rvert$ with $x : S^n \to \lvert X \rvert$ – once we know this, we can apply the hypothesis, and after a bit more messy work, we get the required $\bar{x} : D^{n+1} \to \lvert X \rvert$.
There is also a proof avoiding geometric realisation entirely, but it is equally messy: the idea is exactly the same, but instead of invoking simplicial approximation, we would use subdivision/extension.
