Proving that $S^1$ is not a retract of the disk $D^2$ or of the plane $\mathbb{R}^2$ using a problem that I already solved. 
I've proven that "If $A$ is a retract of $X$ and $X$ is contractible, then $A$ is also contractible."
Can I use this to prove the circle $S^1$ isn't a retract of the disk $D^2$ or of the plane $\Bbb R^2$?If so, how can I do that?

My trial:
$(1)$ I was allowed to take this theorem:

The circle $S^1$ isn' t contractible,

for granted and I can use it. So I decided to prove that the disk $D^2$ (and the plane $\Bbb R^2$ )is contractible (which can be done because any convex space is homotopically equivalent to the space of the one point (by the straight line homotopy and because $D^2$ and $\mathbb{R^2}$ are convex spaces)) and to assume that $S^1$ is a retract of the disk $D^2$ ( or of the plane $\Bbb R^2)$ and then, by the problem I've proven, I should conclude $S^1$ is contractible, which contradicts the theorem  above that I took for granted .... am I correct in this proof?

$(2)$ I have found a solution in this link which uses the idea of the fundamental group and Brouwer fixed point theorem, but, unfortunately, these tools I haven't studied yet.
 A: By contradiction is fine, in detail:


*

*Assume $S^1$ is a retract of $D^2$.

*$D^2$ is contractible (by straight line segments to the origin, as you say).

*Quoted theorem applied to $1$ and $2$: $S^1$ is contractible.

*Fact given to you: $S^1$ is not contractible.


Hence contradiction from $3$ and $4$ and so $1$ must be false: $S^1$ is not a retract of $D^2$.
Another proof could use the Brouwer fixed point theorem:


*

*Assume $S^1$ is a retract of $D^2$.

*Theorem by Brouwer: every $f: D^2 \to D^2$ that is continuous has a fixed point. (i.e. $D^2$ has the FPP (fixed point property)).

*Theorem: if $X$ has the FPP and $A$ is a retract of $X$, then $A$ has the FPP too. (If $g: A \to A$ is continuous, and $r$ is a retraction, apply the FPP to $g \circ r: X \to A \subseteq X$, etc.)

*$1$ to $3$ together imply $S^1$ has the FPP.

*Fact: $S^1$ does not have the FPP (consider the rotation around $90$ degrees either way, e.g.).

*Contradiction between $4$ and $5$ shows $1$ must be false again.

