This is probably not "down-to-Earth" enough for your purposes, but it was one of the first uses of de Rham cohomology that I really enjoyed and I feel like I must share it. (I learned it from Bott's article "The geometry and representation theory of compact Lie groups" in the 1977 Proceedings of the SRC/LMS Research Symposium on Representations of Lie Groups.)
Theorem: If the $n$-sphere $S^n$ is a Lie group, then $n$ must be odd (or zero!).
Proof: $S^0$ is obviously a Lie group, so that's that. If $S^{2n}$ were a Lie group , (and $n>0$), then we'd have a smooth multiplication map $m \colon S^{2n} \times S^{2n} \to S^{2n}$ and an identity element $e \in S^{2n}$. The map $m$ induces a ring homomorphism $m^\ast \colon H^\ast(S^{2n}) \to H^\ast(S^{2n} \times S^{2n})$. Composing this with the Kunneth isomorphism $$ H^\ast(S^{2n} \times S^{2n}) \stackrel{\sim}{\longrightarrow} H^\ast(S^{2n}) \otimes H^\ast(S^{2n}) $$ yields a map $f \colon H^\ast(S^{2n}) \to H^\ast(S^{2n}) \otimes H^\ast(S^{2n})$.
Now,
$$ H^\ast(S^{2n}) = \begin{cases} \mathbb{R} & \text{if }\ast=0 \text{ or } 2n \\ 0 & \text{otherwise.}\end{cases} $$
So, writing $1$ and $\lambda$ for the generators of $H^0$ and $H^{2n}$, respectively, we have $f(1)=1\otimes1$ and $f(\lambda)=a(\lambda\otimes1)+b(1\otimes\lambda)$ for some $a,b\in\mathbb{R}$. To determine $a$ and $b$, first restrict the multiplication map to $S^{2n} \times \{e\}$, where it is the identity map. Consequently $m^*$ gives the identification $H^\ast(S^{2n} \times \{e\}) \cong H^\ast(S^{2n})$ and it follows that $a=1$. Similarly, $b=1$.
As $f$ is a ring homomorphism, we also have
$$\begin{align}
f(\lambda^2) = f(\lambda)^2 &= (\lambda\otimes1 + 1\otimes\lambda)^2 \\
&= \lambda^2\otimes1+ (\lambda\otimes1)(1\otimes\lambda)+(1\otimes\lambda)(\lambda\otimes1)+1\otimes\lambda^2.\end{align}$$
As $\lambda^2 \in H^4 H^{4n} = 0$, this reduces to $0 = (\lambda\otimes1)(1\otimes\lambda)+(1\otimes\lambda)(\lambda\otimes1)$. Recall, however, that the product structure on the tensor product $A \otimes B$ of graded rings is given by
$$ (a \otimes b)(c \otimes d) = (-1)^{\deg a \deg c}(ac\otimes bd). $$
Since $\lambda \in H^2$H^{2n}$, this means that
$$ (\lambda\otimes1)(1\otimes\lambda) = (1\otimes\lambda)(\lambda\otimes1) = \lambda \otimes \lambda.$$
Consequently, $\lambda \otimes \lambda = 0$, which is a contradiction! $\blacksquare$
Of course, there are several other (perhaps even better) ways of proving this theorem, but I think this proof is fairly charming.
