During lecture we defined simply connected set in $\mathbb{R}^n$:
$\Omega \subset \mathbb{R}^n$ is simply connected, iff it is connected and for any $C^1$ closed curve $c:[0,1]\rightarrow \Omega$ there is $C^2$ function $H:[0,1]^2 \rightarrow \Omega$ and point $q$ that $H(t,0)=c(t)$, $H(t,1)=q$ for $t\in [0,1]$.
With this definition I can easily show that any closed 1-form in $\Omega$ is exact. It is straightforward application of Stokes theorem.
What if I define simply connected set in $\mathbb{R}^n$ without $C^1$ and $C^2$ smoothness of curve $c$ and function $H$. I would only require continuousness of $c$ and $H$(this is standard definition of simply connected set). Do I get different classification of simply connected sets?
No, you'll get the same thing. This is basically the statement of smooth homotopy theorem (you can replace smooth with $C^n$ without any change).