At Andre's suggestion, I'll turn my comments into an answer. If we correct the OP by asking about cosimplicial topological spaces, then we have to decide on a notion of equivalence for them, and the obvious one (tying in with DAG and derived differential geometry) is to ask for weak equivalence on the associated simplicial ring of $\mathbb{R}$-valued functions. But then it turns out that the derived structure is meaningless, as everything is equivalent to something with constant cosimplicial structure:

The reason for this is that the ring $A$ of $\mathbb{R}$-valued functions on a topological space has the structure of a $C^0$-ring, meaning that for every continuous function $f : \mathbb{R}^n \to \mathbb{R}$, we have a systematic way of evaluating $f(a_1, \ldots,a_n) \in A$ for all $a_1, \ldots, a_n \in A$. But any simplicial $C^0$-ring is discrete in the sense that $\pi_0A\simeq A$.

This can be proved by looking at the simplicial $C^0$-rings $P^n$ representing $\pi_n$ in the homotopy category. We find that the functor $\pi_n$ is identically $0$ because $\pi_n(P^n)=0$. For the standard cofibrant representative (given by $P^n_i=\mathbb{R}$ for $i<n$, $P^n_n=C^0(\mathbb{R})$, $P^n_{n+1}=C^0(\mathbb{R}^{n+1})$ etc.), this amounts to saying that any function $f$ defined on the line $\langle (1,1, \ldots,1)\rangle \subset \mathbb{R}^{n+1}$ which vanishes at the origin can be extended to a function on $\mathbb{R}^{n+1}$ vanishing on the co-ordinate axes. For continuous functions this can be done, whereas for smooth functions the derivative at $0$ is overdetermined.