I am getting a contradiction by calculating the simplicial DeRham complex of an arbitrary manifold and getting it to be trivial. I also get a similar contradiction using the Godement resolution for topological spaces, but I am stating the DeRham one here. I'd appreciate if anyone can point where the problem is.
Let's assume $\mathcal{F}$ is sheaf on the category of smooth manifolds. For example we will consider the sheaf of $k$-differential forms $\Omega^k$. Let's denote by $C\mathcal{F}$ the sheaf defined in the following way:
$$C\mathcal{F}(X):=\mathcal{F}(\mathbb{R}\times X)$$
We call a sheaf $\mathcal{F}$ to be $\textit{contractible}$ if we have a morphism of presheaves denoted by $p:\mathcal{F}\rightarrow C\mathcal{F}$ such that $i_0\circ p = id$ and $i_1\circ p = 0$ where $i_0$ is the pullback along the closed immersion of $X$ in $X\times \mathbb{R}$ at time $0$ and similarly $i_{1}$ the pullback at time $1$. Now I would argue that the $k$-differential form sheaves are contractible. For $\alpha \in \Omega^k(X)$ define $p(\alpha):=(1-t)\alpha \in \Omega^k(\mathbb{R}\times X)$. It is easy to check $p$ makes the sheaf of differential forms a contractible sheaf.
Let's denote by $\Delta^n$ the $n$-simplex embedded in $\mathbb{R}^n$ i.e. it is just $\mathbb{R}^n$ with faces given by the equations $x_1=0, \ldots x_n =0$ and $\sum_{i=1}^nx_i =0$. If $\mathcal{F}$ is contractible then $\mathcal{F}(X\times \Delta^{\bullet})$ is an acyclic complex. In order to see this we just need to construct a contracting homotopy for the identity map of the complex. This can be done by the map which sends $\beta \in \mathcal{F}(X\times \Delta^n)$ to $p(\beta) \in \mathcal{F}(X\times \Delta^{n+1})$ then we triangulate the image and re-arrange with correct signs. Especially $\Omega^k(X\times \Delta^{\bullet})$ are acyclic for $k\in \mathbb{Z}$. Now let's look at $H^i(X\times \Delta^{\bullet}, \mathbb{R})$ which is the DeRham cohomology of the simplicial manifold $X\times \Delta^{\bullet}$. If we calculate this by replacing $\mathbb{R}$ by its resolution by the DeRham complex and then write the bicomplex spectral sequence there are two ways to filter the bicomplex. One way of it produces a spectral sequence with the first page includes DeRham cohomology $X\times \Delta^n$s which because of homotopy invariance of cohomology gives the cohomology of $X$. The other filtration produces a spectral sequence where the first page consists of the cohomology of the complex $\Omega^p(X\times \Delta^{\bullet})$ which are trivial because of contractibility of differential forms hence giving zero, implying DeRham cohomology is trivial.