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 $\underline{\mathbb R}\to(\Omega_{(-)}^\ast,\operatorname d)$ of chain complexes of sheaves on $M$ is a quasi-isomorphism. This reduces to local charts, which is Poincaré's lemma, and his argument is in some sense akin to what you depicted (but use 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 (this step, you really need the paracompactness).
Combining these two points, we see that the de Rham cohomology is equivalent to the cohomology of the constant sheaf $\underline{\mathbb R}$.