Bounded faces of a planar path 
Let $\gamma:[0,1]\to\mathbb{R}^2$ be a path.  Show that for two disjoint compact sets $K_1$ and $K_2$ in $\mathbb{R}^2$, there are only finitely many components of $\gamma^c$ which intersect both $K_1$ and $K_2$.

Since $\gamma$ is continuous on a compact set it is uniformly continuous.  In contrapositive form, this says that for any $M>0$, there is a $\delta>0$ such that if $|f(s)-f(t)|>M$, then $|s-t|>\delta$.  I believe that the result should follow from this fact.
NOTE: I posted a solution below for the special case where $K_1$ and $K_2$ are circles.  It seems much longer than necessary, I think there should be a short proof of the general result.
 A: Following is a solution for the case that $K_1$ and $K_2$ are circles.
For $r>0$, define $C_r$ to be the circle centered at the origin, with radius $r$.  By first applying a homeomorphism to the sphere, we may assume without loss of generality that $K_1=C_1$ and $K_2=C_2$.
Suppose by way of contradiction that there are infinitely many components of $\gamma^c$ which intersect both $C_1$ and $C_2$.  Since each component of $\gamma^c$ is open, there are only countably many of them.  Let $\{E_i\}_{i=1}^\infty$ be an enumeration of the components of $\gamma^c$ which intersect both $C_1$ and $C_2$.  Since each $E_i$ is open and connected in $\mathbb{C}$, it is also path connected.  Therefore for each $i$, we may define a path $\varphi_i:[0,1]\to E_i$ such that $\varphi_i(0)\in C_1$ and $\varphi_i(1)\in C_2$.  By restricting $\varphi_i$ if necessary, we may also assume that for each $t\in(0,1)$, $|\varphi_i(t)|\in(1,2)$.  Define $x_i=\varphi_i(0)$ and $z_i=\varphi_i(1)$.  Define $r_i\in(0,1)$ to be the smallest number such that $|\varphi_i(r_i)|=1.5$, and define $y_i=\varphi_i(r_i)$.
It is immediate from their definition that, for any $i,j\in\mathbb{N}$, if $i\neq j$, then $\varphi_i$ and $\varphi_j$ are distinct from each other.  Since they do not cross in the region $\{z:1<|z|<2\}$, it follows that the orientation of the $x_i$ around $C_1$ is the same as the orientation of the $z_i$ around $C_2$ and the $y_i$ around $C_{1.5}$.  That is, for any distinct $i,j,k\in\mathbb{N}$, if $x_i,x_j,x_k$ is the order in which the points appear when $C_1$ is traversed with positive orientation, then $z_i,z_j,z_k$ is the order in which those points appear when $C_2$ is traversed with positive orientation, and $y_i,y_j,y_k$ is the order in which those points appear when $C_{1.5}$ is traversed with positive orientation.
Since $\gamma$ is continuous on the compact set $[0,1]$, it is uniformly continuous, therefore we may choose a $\delta>0$ such that for any $s,t\in[0,1]$, if $|\gamma(s)-\gamma(t)|>1/4$, $|s-t|>\delta$.  Choose an $N>0$ such that $N\delta>1$.  By reordering the elements described above, we may assume that the points $x_1,x_2,\ldots,x_N$ appear in precisely this order as $C_1$ is traversed with positive orientation.  For the remainder of the proof, addition is done modulo $N$.
For each $i$, let $F_i$ denote the domain which is bounded by the paths $\varphi_i$, $\varphi_{i+1}$, the arc of the circle $C_1$ with end points $x_i$ and $x_{i+1}$ which does not contain $x_{i+2}$, and the arc of the circle $C_2$ with end points $z_i$ and $z_{i+1}$ which does not contain $z_{i+2}$.  Since $y_i$ and $y_{i+1}$ are in different faces of $\gamma$, the arc of $C_{1.5}$ with end points $y_i$ and $y_{i+1}$ which does not contain $y_{i+2}$, must intersect $\gamma$.  Since this arc is contained in $F_i$, we may choose a point $\alpha_i\in\gamma\cap F_i$ such that $|\alpha_i|=1.5$.  Define $s_i\in(0,1)$ to be the smallest number such that $\gamma(s_i)=\alpha_i$.  Define $t_i\in(s_i,1)$ to be the smallest number (if one exists) such that $\gamma(t_i)\in\partial F_i$ (if no such number exists, choose $t_i\in(0,s_i)$ to be the largest such that $\gamma(t_i)\in\partial F_i$).  Define $\beta_i=\gamma(t_i)$.  Let $J_i$ denote the interval in $[0,1]$ with end points $s_i$ and $t_i$ (that is, either $[s_i,t_i]$ or $[t_i,s_i]$).
The point $\beta_i\in\partial F_i$ must be in either $C_1$ or $C_2$, since the portion of $\partial F_i$ which is not in either $C_1$ or $C_2$ (namely $\varphi_i$ and $\varphi_{i+1}$) is contained in $\gamma^c$.  Thus it follows that $|\gamma(s_i)-\gamma(t_i)|>1/4$, and thus $\lambda(J_i)>\delta$.  Note that for any distinct $i,j\in\{1,2,\ldots,N\}$, since $F_i$ and $F_j$ are disjoint, $\gamma(J_i)$ and $\gamma(J_j)$ are disjoint, and thus $J_i$ and $J_j$ are disjoint.  We thus have that $\{J_i\}$ is a sequence of disjoint intervals in $[0,1]$, giving us the contradiction $$1=\lambda([0,1])\geq\sum_{i=1}^N\lambda(J_i)\geq\sum_{i=1}^N\delta=N\delta>1.$$
This concludes the proof.
A: Here is a reduction of the general case to the one of two disjoint round circles (which you already know how to handle). Suppose that there exists a 
sequence $U_i$ of distinct components of $R^2 - Im(\gamma)$ each of which intersects both $K_1$ and $K_2$ nontrivially. Then, after passing to a subsequence (twice), by compactness of $K_1, K_2$, there exist points $x_j\in K_j, j=1,2$ such that 
$$
\lim_{i\to\infty} dist(x_j, cl(U_i))= 0, j=1,2$$
Now, since $x_1\ne x_2$,   there is a pair of disjoint closed round disks $B_1=B(x_1, r), B_2=B(x_2, r)$ such that each $U_i$ has nonempty intersection with both $B_1$ and $B_2$. By connectedness of $U_i$'s, each $U_i$ has nonempty intersection with each round circle $C_j=\partial B_j, j=1,2$. 
