To show that a space of contact forms in $\mathbb{R}^3$ is contractible, it's enough to show that $$\lambda_t=(1-t)\lambda_1+t\lambda_2$$ is a contact form for all $t\in[0,1]$ if $\lambda_0,\lambda_1$ are contact forms.
By the definition, $\lambda$ is a contact form if $\lambda\wedge d\lambda$ is a volume form on $\mathbb{R}^3$ i.e. $\lambda\wedge d\lambda\neq 0$. Then observe that $$\lambda_t\wedge d\lambda_t=(1-t)^2\lambda_1\wedge d\lambda_1+t^2\lambda_2\wedge d\lambda_2+$$ $$+t(1-t)\lambda_1\wedge d\lambda_2+t(1-t)\lambda_2\wedge d\lambda_1.$$ Assume that $\lambda_t\wedge d\lambda_t=0$ i.e. there exists a non-zero $x\in \mathbb{R}^3$ such that for all $v,w\in \mathbb{R}^3$, $$(\lambda_t\wedge d\lambda_t)(x,v,w)=0.$$ The main thing that I want to argue that we can choose $v$ and $w$ in $\mathbb{R}^3$ such that $d\lambda_1(v,w)=0$, $d\lambda_2(v,w)\neq0$, and $v\in \ker(\lambda_1)$. Then we have that $$0=(\lambda_t\wedge d\lambda_t)(x,v,w)=(t^2\lambda_2\wedge d\lambda_2+t(1-t)\lambda_1\wedge d\lambda_2)(x,v,w)=(t^2\lambda_2\wedge d\lambda_2)(x,v,w)\neq 0$$ which gives us a contradiction. However, how can I gurantee that $x\not\in\ker(\lambda_2)$? So, if I can find such $v,w$ and $x\not\in\ker(\lambda_2)$, then $\lambda_t\wedge d\lambda_t$ is indeed a contact form on $\mathbb{R}^3$.
I hope this proof more or less correct. If I want to generalize it to $\mathbb{R}^{2n+1}$, do I work in the same way? Or, is there more consice intuitive proof of this fact?