The boundary complex of the $5$-simplex is a simplicial $4$-sphere

Let $L$ be the abstract simplicial complex on the six vertices $\{0,\dots,5\}$ whose faces are exactly the nonempty proper subsets of the vertex set, i.e. the boundary complex $\partial\Delta^5$ of the $5$-simplex. Then the barycentric-coordinate geometric realization of $L$ is homeomorphic to the unit sphere

so $L$ is a simplicial $4$-sphere.

Concretely, $|L|$ is the set of probability vectors $x\in\mathbb R^6$ (nonnegative coordinates summing to $1$) at least one of whose coordinates vanishes, that is, the boundary of the standard $5$-dimensional simplex. Radial projection from the barycentre $(1/6,\dots,1/6)$ within the hyperplane $\{\sum_i y_i=0\}$, which is $5$-dimensional, identifies this boundary with a $4$-sphere: every ray leaving the barycentre meets the boundary exactly once, and the realization is compact, so the resulting continuous bijection is a homeomorphism.

This provides a concrete simplicial $4$-sphere in the sense used by the mission, the base case for constructions of triangulations of $S^4$ with prescribed full subcomplexes.

Formalization Note The complex is pinned down by the hypothesis that a finite set of vertices is a face if and only if it is nonempty and not the whole vertex set; the conclusion is the existence of a homeomorphism between the realization and the Euclidean unit $4$-sphere.