In my manifolds course, we proved that if two morphisms $\phi,\psi:M \to N$ are homotopic, then the induced pullbacks $\phi^*,\psi^*:H^r(N) \to H^r(M)$ on the de Rham Cohomologies are equal. We then used the fact that $\phi = id_{\mathbb{R}^n}$ and $\psi = 0 \in \mathbb{R}^n$ are homotopic and that their pullbacks are the identity and $0$ map respectively to show that this to show that $H^r(\mathbb(R^n)) = \{0\}$ for $r>0$, as this is the only VS where the identity map and zero map are equal. However, I don't see what exactly goes wrong for $r=0$, which is not $\{0\}$, but $\mathbb{R}$.
My guess is that somehow the pullback of $\psi$ is not actually the $0$ map on $H^r(\mathbb{R}^n)$, but I don't know why this would be true or indeed true at all.
Let $\alpha$ be a zero form on $M$, that is, $\alpha$ is a function. Let $\psi : N \to M$ be a constant map sending $\psi (n)$ to $m_0$ for all $n\in N$. Then $\psi^* \alpha (n) = \alpha (m_0)$ for all $n\in N$ and $\psi^* \alpha$ is a constant zero form which might or might not be zero.
The main difference is that for $r >0$, the definition of $\psi^*$ involves derivatives of $\psi$, but not for $r=0$.