Continuous map homotopic to Id in infinite tangent circles space

Consider the space of infinite tangent circles $$C= \bigcup_{n=1}^\infty S( (0, \tfrac{1}{n}),\tfrac{1}{n}) ,$$ where $S(x,r)$ is the sphere on $\mathbb{R}^2$ with center $x$ and radius $r$ (see the picture below). On $C$ we consider the subspace topology induced from $\mathbb R ^2$.

Show that every continuous map $\varphi : C \to C$ homotopic to the identity holds $$\varphi (0,0)=(0,0).$$

Remark. I have been suggested to use that none of the inclusions of the circles into $C$ is homotopic to a constant map.

My attemp: I tried to show that if $\varphi (0,0) = x_0 \in S( (0, \tfrac{1}{n}),\tfrac{1}{n})$ for some $n$, then $\varphi_{| S( (0, \tfrac{1}{n}),\tfrac{1}{n})}$ is homotopic to the constant map $x_0$, but I don't really know how to finish the argument. • Should the homotopy fix the origin, i.e. $\phi_t(0,0)=(0,0)$ for all $t$? – klirk Nov 25 '17 at 17:15
• @klirk I obtain that as a consequence of the statement, but that hypothesis is not assumed before – Minkowski Nov 25 '17 at 17:30
• I ask, because the first thing i could think about was if we consider for the homotopy $H$: $H( (0,0),\cdot)$, then this would give a path from the origin to a distinct point. But as $C$ is path connected, this is not really useful. – klirk Nov 25 '17 at 17:59

Let $$S_n = S((0, \tfrac{1}{n}),\tfrac{1}{n})$$ and $$p = (0,0)$$. Let $$i_n : S_n \to C$$ denote inclusion. We define $$r_n : C \to S_n, r_n(x) = x$$ for $$x \in S_n$$ and $$r_n(x) = p$$ for $$x \in S_m$$ with $$m \ne n$$. This map is a continuous retraction.
Let $$\varphi : C \to C$$ be a map which is homotopic to the identity. Then also all $$\psi_n = r_n \circ \varphi \circ i_n : S_n \to S_n$$ are homotopic to $$r_n \circ id_C \circ i_n = r_n \circ i_n = id_{S_n}$$. In particular these maps are not null homotopic since the $$S_n$$ are homeomorphic to the circle $$S^1$$.
The sets $$T_n = S_n \setminus \{p\}$$ are all open in $$C$$ and homeomorphic to $$(0,1)$$ (and therefore contractible).
Assume $$\varphi(p) \ne p$$. Then $$\varphi(p) \in T_n$$ form some $$n$$. By continuity there exists an open neighborhood $$U$$ of $$p$$ in $$C$$ such that $$\varphi(U) \subset T_n$$. Clearly $$U$$ contains all but finitely many $$S_m$$. So take any $$m$$ such that $$S_m \subset U$$. But then $$\psi_m$$ is null homotopic since it factors through $$\varphi \circ i_m$$ which is null homotopic because its image is contained in the contractible $$T_n$$. This is a contradiction. Therefore $$\varphi(p) = p$$.
If $\varphi$ is homotopic to de identity then it induces an application $\varphi_*:\pi_1(C)\rightarrow\pi_1(C)$ that equals the identity. Then obviously $\varphi$ cannot send one circumference (actually, element of the fundamental group) to any other, so if $S$ is any circumferemce $S\subseteq\varphi(S)$ (*). This means that for any circumference $S_m$, with the obvious notation, we have a point $x_m$ such that $\varphi(x_m)\in S_m$. $x_m$ tends to $(0,0)$ so if $\varphi(0,0)=p\neq (0,0)$ then $\varphi(x_m)$ tends to $p$, but this is absurd as $p\neq 0$.
(*)This can be proven directly without using the fundamental group. Try to push the center of an "8" to one of the circumferences but keeping the shape of "8", the same argument can be used to $C$.