I would not believe that the paracompactness is avoidable. In any case, Dustin Clausen's Lectures on algebraic de Rham cohomology, in particular, Lecture 1, should cover some of ingredients that you want. The argument works as follows:
Let $M$ be a manifold.
The map $\underline{\mathbb R^{\operatorname{disc}}}\to(\Omega_{(-)}^\ast,\operatorname d)$ of cochain complexes of sheaves on $M$, where the constant sheaf $\underline{\mathbb R^{\operatorname{disc}}}$ is viewed as a cochain complex concentrated in degree 0, is a quasi-isomorphism. This does not need paracompactness, and it reduces to local charts, which is Poincaré's lemma, and his argument is in some sense akin to what you depicted (but uses the projective tensor product of Fréchet spaces instead of the algebraic one — surely the algebraic one is not the "correct" one in this case).
For every $r\in\mathbb N$, the sheaf $\Omega_{(-)}^r$ has vanishing higher cohomology, since it is a $C^\infty(-)$-module (for this step, you really need the paracompactness).
Combining these two points, and apply the de Rham–Weil theorem, we see that the de Rham cohomology, as an object in the derived category $D(M)$ of abelian sheaves, is equivalent to the cohomology $R\Gamma(M;\mathbb R)$ of the constant sheaf $\underline{\mathbb R^{\operatorname{disc}}}$.