The following is essentially a fleshed-out version of the proof of Proposition 4.2 in Ghys's beautiful article Groups acting on the circle.
It is convenient to translate the question into a question about continuous functions on $\mathbb{R}$. Viewing the circle $S^1 = \mathbb{R}/\mathbb{Z}$ as a quotient of $\mathbb{R}$, a continuous function $f\colon S^1 \to S^1$ can be lifted over the covering projection $\pi \colon \mathbb{R} \to S^1$ to a continuous function $F \colon \mathbb{R} \to \mathbb{R}$ such that $f \circ \pi = \pi \circ F$: on the fundamental domain $[0,1)$ the function $F$ is determined by $f$ up to addition of an integer, and once that integer is chosen, there is only one way to extend $F$ continuously to all of $\mathbb{R}$.
A continuous function $F \colon \mathbb{R} \to \mathbb{R}$ is the lift of a continuous function $f \colon S^1 \to S^1$ if and only if there is $d \in \mathbb{Z}$ such that for all $x$ the equation $F(x+1) = F(x) + d$ holds. It is not hard to see that $d = \deg{f}$. Using this [or by elementary considerations with the intermediate value theorem] we see that $f$ can only be a homeomorphism if $d = \pm 1$: the lifts of a homeomorphism must satisfy $F(x+1) = F(x) \pm 1$. The lift $F$ of a homeomorphism $f$ is a homeomorphism because continuous bijections $\mathbb{R} \to \mathbb{R}$ have continuous inverses.
Let $\tilde{H} = \operatorname{Homeo}_\mathbb{Z}(\mathbb{R})$ be the group of homeomorphisms $F \colon \mathbb{R} \to \mathbb{R}$ such that $F(x+1) = F(x) +1$.
The compact-open topology on $\tilde{H}$ coincides with the uniform topology and is metrized by $d(F,G) = \sup_{x \in [0,1]} \lvert F(x) - G(x)\rvert$ since $F-G$ is $1$-periodic. If $F,G \in \tilde{H}$ then their convex combinations $(1-t)F + tG$ belong to $\tilde{H}$ for all $t \in [0,1]$: they are strictly increasing and surjective. The map $$h\colon [0,1] \times \tilde{H} \times \tilde{H} \longrightarrow \tilde{H}, \quad(t,F,G) \longmapsto (1-t)F + tG$$ is clearly continuous. This shows that $\tilde{H}$ is contractible, as we can take $G$ to be the identity $G(x) = x$ and for all $F \in \tilde{H}$ we have $h(0,F,G) = F$ and $h(1,F,G) = G$.
The map $F \mapsto f$ sending $F \in \tilde{H}$ to the homeomorphism of $S^1$ it covers is a continuous homomorphism $p\colon \tilde{H} \to H = \operatorname{Homeo}_+(S^1)$ onto the group of orientation-preserving homeomorphisms (i.e., those $f \colon S^1 \to S^1$ preserving the cyclic order of triples of pairwise distinct points on the circle). The kernel of $p$ can be identified with $\mathbb{Z} = \langle \tau \rangle$ generated by the unit translation $\tau(x) = x + 1$. In fact, $p$ identifies $\tilde{H}$ with the universal covering group of $H$.
Observe that $f = p(F)$ is a rotation with angle $[\alpha] \in \mathbb{R/Z}$ if and only if $F(x) = x + \alpha$ is a translation.
Now notice that each $F \in \tilde{H}$ can be uniquely written as $F(x) = x + \alpha_F + \varphi_F(x)$ with $\alpha_F = \int_{0}^1 [F(x)-x]\,dx \in \mathbb{R}$ and $\varphi(x) = F(x) - x -\alpha$ is a $1$-periodic function with mean zero: $\int_{0}^1 \varphi(t)\,dt = 0$. Since $F_n \to F$ in $\tilde{H}$ means uniform convergence, we see that $\alpha_F$ and $\varphi_F$ depend continuously on $F$. Thus, the map
$$
\tilde{k} \colon [0,1] \times \tilde{H} \longrightarrow \tilde{H} \quad
\tilde{k}(t,F) = x + \alpha_F + (1-t)\varphi_F
$$
is continuous and $\tilde{k}(0,F) = F$ while $\tilde{k}(1,F) = x+\alpha_F$ covers a rotation, i.e., an element of $SO(2)$. If $F(x) = x + \alpha$ is a translation then $\tilde k(t,F) = F$ for all $t \in [0,1]$, so $\tilde k$ is constant on the group of translations.
For two lifts $F_1(x) = x + \alpha_{F_1} + \varphi_{F_1}(x)$ and $F_{2}(x) = x + \alpha_{F_2} + \varphi_{F_2}(x)$ of the homeomorphism $f \colon S^1 \to S^1$ we have $\varphi_{F_1} = \varphi_{F_2}$ and $\alpha_{F_1} - \alpha_{F_2} \in \mathbb{Z}$. This implies that $\tilde{k}(t,F_1)$ and $\tilde{k}(t,F_2)$ cover the same homeomorphism of $S^1$ and hence $\tilde{k}$ yields a continuous map $k\colon [0,1] \times H \to H$ which is a deformation retraction from $\operatorname{Homeo}_+(S^1)$ to $SO(2)$.
The argument for the group of all homeomorphisms of $S^1$ is essentially the same: If $f$ reverses orientation then its lifts are of the form $F(x+1) = F(x) - 1$ and $F$ is strictly decreasing. Now define $\alpha_F = \int_{0}^1 [F(x) + x]\,dx$ and $\varphi_F = F + x - \alpha_F$ to obtain a homotopy from $F$ to $\alpha_F - x$, covering the reflection at the line with angle $\alpha_F/2$.
You want to show that $f(t, x) = (1 - t)x + tx/|x|$ is not in $A$ for any $t\in [0, 1]$ and any $x\notin A$.
Suppose it is not the case. Then there is $x\notin A$ and $t\in [0, 1]$ such that $f(t, x) = \lambda x \in A$, where $\lambda\in \Bbb R$ is the number $1 - t + t/|x|$.
If $|x| > 1$, then we have $\lambda \geq (1 - t)/|x| + t/|x| = 1/|x|$, hence $|f(t, x)| \geq 1$. This contradicts the assumption that every point in $A$ has norm $< 1$.
If $|x| \leq 1$, then we have $\lambda \geq 1 - t + t = 1$. But we also have $0\in A$. By convexity of $A$, the linear combination $1/\lambda \cdot f(t, x) + (1 - 1/\lambda) \cdot 0 = x$ is in $A$. This contradicts the assumption $x\notin A$.
Best Answer
Say the circle has radius $1$, center at the origin, and lies in the $xz$-plane. Now let $S^2$ be embedded with center in the origin and radius $2$. Everything outside the sphere deformation retracts onto the sphere.
For every point $x$ inside the sphere, if it's on the $y$-axis, leave it. Otherwise, there is a point $p$ on the circle that is closest to it. Now let $x$ move directly away from $p$ until it either hits the $y$-axis or the sphere, whichever comes first (this is continuous because the line between $p$ and $x$ is coplanar to the $y$-axis, so there are no points that "just misses" the $y$-axis and goes past it). You've now created the second space.