Diff$(S^1)$ is deformation retracts to $O(2)$ I have proved the following Diff$^+(S^1$) is path connected. Now I want to prove it is deformation retracts to $O(2)$. 
What I tried is the following:
I define an onto homomorphism $$ f:\text{Diff}^{+}(\mathbb{D}^2)\to \text{Diff}^{+}(S^1) $$ by $$ \phi\mapsto \phi|_{\partial \mathbb{D}^2=S^1}$$
This is onto because any diffeomorphism of $S^1$ can be extended to a diffeomorphism of $\mathbb{D^2}$. The kernel of this homomorphism is $\text{Diff}^+(\mathbb{D}^2\text{rel } \partial)$. After that I don't know what to do. 
 A: We will show the result for homotopy equivalences of $S^1$, denoted $Self(S^1)$, and then show that we can  restrict down to either the homeomorphism or diffeomorphism group. This is all discussed in Lurie's notes, but I'll reproduce the argument for completeness.
both $Diff(S^1)$ and $Self(S^1)$ can be decomposed into two pieces, orientation preserving and orientation reversing, and likewise so can $O(2)$.
Hence, it is enough to show that $SO(2)$ is a deformation retract of $Diff^+(S^1)$, as noted in the comments.
First, note that there are decompositions $Diff^+(S^1)=Diff_0^+(S^1)SO(2)$ and $Self^+=Self_0^+(S^1)SO(2)$, where the subscript $0$ indicates that they fix a base point of $S^1$. In other words, every diffeomorphism can be obtained from one that fixes a base point, and then rotates appropriately.
First, we can regard $S^1= \mathbb R/\mathbf Z$, so a base-point perserving map $f:S^1 \to S^1$ can be regarded as a map $\tilde{f}:\mathbb R \to \mathbb R$ where $f(x+1)=f(x) \pm d$ and $\tilde{f}(0)=0$ (take the perserved fixed point to be the image of $0$.) Note that $f$ is a homotopy equivalence if and only if $d= \pm 1$, since $d$ is the degree of the map, and since it is orientation perserving, $d=1$.
Hence, we have an identification $Self_0^+(S^1)=\{\tilde{f}:\mathbb R \to \mathbb R \mid \tilde{f}(0)=0, f(x+1)= x+1\}$.
However, this lets us form the straight line homotopy $F_t:=(1-t)\tilde{f}(x)+tx$, which shows the space to be contractible (retractible)  (as all functions are homotopic to identity)
$Diff_0^+(S^1) \subset Self_0^+(S^1)$ in the above identification by taking the subset of smooth maps without vanishing derivative. We can follow the same homotopy.
In particular, this shows a retract via the straight line homotopy from $Diff^+(S^1)$ to $SO(2)$
