If two closed plane curves are outside each other, can there be a point inside both of them? I think this recent
question (also here)
has a quick answer if the conjecture below is true.
It looks "obviously" true, but I've learned to distrust my judgement
in such matters.  It also looks as if it should have an "obvious" proof,
but I'm having no ideas.
Definitions
Let $[a, b]$ be any compact interval in $\mathbb{R}$. For any
continuous function $\gamma \colon [a, b] \to \mathbb{C}$, denote
the connected, compact set of points $\{\gamma(t) : a \leqslant t \leqslant b\}$
by $[\gamma]$. Define $r(t) = |\gamma(t)|$ ($a \leqslant t \leqslant b$).
By Theorem 7.2.1 of A. F. Beardon, Complex Analysis (Wiley, Chichester 1979),
there exists a branch of $\operatorname{Arg}\gamma$ on $[a, b]$, i.e.
a continuous function $\theta \colon [a, b] \to \mathbb{R}$ such that
$\gamma(t) = r(t)e^{i\theta(t)}$ ($a \leqslant t \leqslant b$).
Quoting from the same reference:

Definition 7.2.1
  Let $\gamma \colon [a, b] \to \mathbb{C}$ be any curve and suppose
  that $w \notin [\gamma]$. We define the index $n(\gamma, w)$ of
  $\gamma$ about $w$ by
  $$
n(\gamma, w) = \frac{\theta(b) - \theta(a)}{2\pi},
$$
  where $\theta$ is any branch of $\operatorname{Arg}(\gamma - w)$ on
  $[a, b]$. If $\gamma$ is closed then $n(\gamma, w)$ is an integer.
The index $n(\gamma, w)$ is sometimes called the winding number
  of $\gamma$ about $w$, for it represents the number of times
  that a point $z$ moves around $w$ as it moves from $\gamma(a)$ to
  $\gamma(b)$ along $\gamma$. [...]
The index can be used to clarify the difficult question of what is
  meant by the 'inside' and 'outside' of a closed curve $\gamma$. We
  shall say
(a) that $z$ is inside $\gamma$ if $z \notin [\gamma]$ and
  $n(\gamma, z) \ne 0$,
(b) that $z$ is on $\gamma$ if $z \in [\gamma]$, and
(c) that $z$ is outside $\gamma$ if $z \notin [\gamma]$ and
  $n(\gamma, z) = 0$.
[...] Observe that [...] the outside of $\gamma$, say $O(\gamma)$,
  is the union of those components of $\mathbb{C} \setminus [\gamma]$
  on which the index is zero.  Thus $O(\gamma)$ is an open set.
  Further [...] $O(\gamma)$ contains the complement of some closed disc.
  If we denote the inside of $\gamma$ by $I(\gamma)$, then
  $$
\mathbb{C} \setminus O(\gamma) = [\gamma] \cup I(\gamma),
$$
and so the set of points which lie inside or on $\gamma$ is a compact set.

Conjecture

For closed curves $\sigma$ and $\tau$,
  if $[\sigma] \subset O(\tau)$ and $[\tau] \subset O(\sigma)$, then
  $I(\sigma) \subset O(\tau)$.

Because the premise $[\tau] \subset O(\sigma)$ implies
$I(\sigma) \cap [\tau] = \emptyset$, the conclusion can be expressed
symmetrically as $I(\sigma) \cap I(\tau) = \emptyset$ - whence the
title of the question.
 A: (Thanks to Moishe Cohen for his very helpful hint. He is not to blame for the use I have made of it, however!)

Lemma If $E$ is a connected, closed subset of a connected, normal topological space $X$, and $G$ is a connected component of $X \setminus E$, then $G \cup E$ is connected.

Proof Define $H = \overline{G} \cap E$.
If $H = \emptyset$, then $\overline{G}, E$ are disjoint closed subsets of $X$, whence there exist disjoint open subsets $U, V$ of $X$ such that $\overline{G} \subseteq U$ and $E \subseteq V$. We cannot have $U = \overline{G}$, because then $U$ would be a closed, open, non-empty proper subset of $X$, so $X$ would be disconnected, contrary to hypothesis. On the other hand, we cannot have $U \ne \overline{G}$, because then $G$ would be properly contained in the open subset $U$ of $X \setminus E$, contrary to its definition.
Therefore, $H \ne \emptyset$.
We have $G \subseteq G \cup H \subseteq \overline{G}$, therefore $G \cup H$ is connected.
Also, $(G \cup H) \cap E = H \ne \emptyset$.
Since $G \cup H$ and $E$ are both connected, it follows that their union $(G \cup H) \cup E = G \cup E$ is connected. $\square$

Corollary The union of $E$ with any collection of connected components of $X \setminus E$ is connected. $\square$
Proposition If $\sigma$ is a closed curve in $\mathbb{C}$, then $[\sigma] \cup I(\sigma)$ is connected.

Proof Let $z, w \in \mathbb{C}$. If $z \in I(\sigma)$, then by definition $n(\sigma, z) \ne 0$. If $z, w$ are in the same connected component of $\mathbb{C} \setminus [\sigma]$, then $n(\sigma, w) = n(\sigma, z) \ne 0$, whence $w \in I(\sigma)$. Therefore, $I(\sigma)$ is a union of connected components of $\mathbb{C} \setminus [\sigma]$. The result now follows from the above corollary. $\square$

Theorem For closed curves $\sigma$ and $\tau$ in $\mathbb{C}$,
if $[\sigma] \subset O(\tau)$ and $[\tau] \subset O(\sigma)$ then
$I(\sigma) \subset O(\tau)$.

Proof Because $[\tau] \subset O(\sigma)$, we have $I(\sigma) \subseteq \mathbb{C} \setminus [\tau]$, therefore $[\sigma] \cup I(\sigma) \subseteq \mathbb{C} \setminus [\tau]$.
It follows that the connected set $[\sigma] \cup I(\sigma)$ is contained in the same
connected component of $\mathbb{C} \setminus [\tau]$ as $[\sigma]$. By hypothesis, this
component is a subset of $O(\tau)$, whence $I(\sigma) \subset O(\tau)$. $\square$
