Here is a quick and easy proof - thanks in good part to a helpful
hint given by Moishe Cohen, in a comment on my recent question cited
below. (I repeat here the quotation included in that question, to
make this answer self-contained, although unfortunately this makes
the proof appear longer than it is!)
Let $[a, b]$ be a compact interval in $\mathbb{R}$. For any
continuous function $\gamma \colon [a, b] \to \mathbb{C}$, denote
the compact, connected set of points $\gamma([a, b])$ by $[\gamma]$,
and define:
\begin{align*}
i\gamma \colon [a, b] \to \mathbb{C}, \ & t \mapsto i(\gamma(t)), \\
-\gamma \colon [a, b] \to \mathbb{C}, \ & t \mapsto -(\gamma(t)), \\
-i\gamma \colon [a, b] \to \mathbb{C},\ & t \mapsto -i(\gamma(t)).
\end{align*}
If $0 \in [\gamma]$, then also $0 \in i[\gamma] = [i\gamma]$, so
the curves $\gamma, i\gamma$ intersect. We assume from now on that
$0 \notin [\gamma]$.
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:
$$
\gamma(t) = r(t)e^{i\theta(t)} \quad (a \leqslant t \leqslant b).
$$
The functions $\theta + \frac{\pi}{2}$, $\theta + \pi$,
$\theta - \frac{\pi}{2}$ are branches of
$\operatorname{Arg}(i\gamma)$, $\operatorname{Arg}(-\gamma)$,
$\operatorname{Arg}(-i\gamma)$, respectively, on $[a, b]$.
Quoting from the same book:
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.
Quoting now from D. J. H. Garling, A Course in Mathematical Analysis, vol. III
(Cambridge 2014) - the slight clash of notation should cause no confusion -
If $\gamma \colon [a, b] \to X$ and $\delta \colon [c, d] \to X$ are
[curves], and $\gamma(b) = \delta(c)$, the juxtaposition
$\gamma \vee \delta$ is the [curve] from $[a, b + (d - c)]$ into $X$
defined by $(\gamma \vee \delta)(x) = \gamma(x)$ for $x \in [a, b]$
and $(\gamma \vee \delta)(x) = \delta(x + (c - b))$ for
$x \in [b, b + (d - c)]$. [...]
Suppose that $\gamma \colon [a, b] \to X$ is a [curve], and that
$w \notin [\gamma]$. [...] If $\gamma = \alpha \vee \beta$ is the
juxtaposition of two [curves] then
$$
n(\gamma, w) = n(\alpha, w) + n(\beta, w).
$$
Using the hypothesis that $\gamma(b) = -\gamma(a)$, we form the
juxtapositions
\begin{gather*}
\sigma = \gamma \vee (-\gamma) \colon [a, 2b - a] \to \mathbb{C}, \\
\tau = (i\gamma) \vee (-i\gamma) \colon [a, 2b - a] \to \mathbb{C},
\end{gather*}
and observe that, by the same hypothesis, these are closed curves.
Clearly $\tau = i\sigma$, i.e. $\tau(t) = i(\sigma(t))$
($a \leqslant t \leqslant 2b-a$).
Using the hypothesis $\gamma(b) = -\gamma(a)$ for a third time,
we have:
$$
n(\gamma, 0) = n(-\gamma, 0) = n(i\gamma, 0) = n(-i\gamma, 0) =
m + \tfrac{1}{2}, \text{ for some } m \in \mathbb{Z},
$$
and consequently:
$$
n(\sigma, 0) = n(\tau, 0) = 2m + 1.
$$
All we need retain from this are the implications
$n(\sigma, 0) \ne 0$, $n(\tau, 0) \ne 0$, i.e.,
\begin{equation}
\label{3109299:eq:1}\tag{1}
0 \in I(\sigma) \cap I(\tau).
\end{equation}
Define $r(t) = |\gamma(t)| = |i\gamma(t)|$
($a \leqslant t \leqslant b$). Being a continuous function on a
compact set, $r$ attains a maximum value, $r(t_0)$. Extending $r$
continuously to the interval $[a, 2b-a]$, by writing
$r(t) = r(t + a - b)$ ($b \leqslant t \leqslant 2b -a$), we see that
$r(t_0)$ is also the maximum value of $|\sigma(t)| = |\tau(t)|$
($a \leqslant t \leqslant 2b-a$).
Suppose that $[\sigma], [\tau]$ are disjoint. A straight line
segment connects the point $\sigma(t_0) = \gamma(t_0)$ to the point
$2\gamma(t_0)$, and it lies entirely in the complement
$\mathbb{C} \setminus [\tau]$. As the second
point clearly lies in the unbounded component of
$\mathbb{C} \setminus [\tau]$, so must the first; therefore, the
whole of the connected set $[\sigma]$ lies in the unbounded
component of $\mathbb{C} \setminus [\tau]$. Similarly, $[\tau]$ lies
in the unbounded component of $\mathbb{C} \setminus [\sigma]$.
All we need retain from this are the implications:
\begin{equation}
\label{3109299:eq:2}\tag{2}
[\sigma] \subset O(\tau) \text{ and } [\tau] \subset O(\sigma).
\end{equation}
That \eqref{3109299:eq:1} and \eqref{3109299:eq:2} stand in
contradiction to one another (for any closed plane curves
$\sigma, \tau$, and with an arbitrary point of the plane
in place of $0$) was precisely the content of the conjecture in my
question
yesterday. I can now proudly report that my weedy little conjecture
has grown up into a beefy big theorem! Thus, the supposition that
$[\sigma], [\tau]$ are disjoint must be false.
For $S \subseteq \mathbb{C}$, let $-S, iS, -iS$ denote the sets
$\{-s : s \in S\}$, $\{is : s \in S\}$, $\{-is : s \in S\}$,
respectively. We have just established that the sets
\begin{gather*}
[\sigma] = [\gamma] \cup [-\gamma] = [\gamma] \cup -[\gamma], \\
[\tau] = [i\gamma] \cup [-i\gamma] = i[\gamma] \cup -i[\gamma]
\end{gather*}
intersect. Let $z$ be a point common to both.
If $z \in [\gamma] \cap [i\gamma]$, we are done right away.
If $z \in [-\gamma] \cap [-i\gamma]$, then
$-z \in [\gamma] \cap [i\gamma]$.
If $z \in [\gamma] \cap [-i\gamma]$, then
$iz \in [\gamma] \cap [i\gamma]$.
If $z \in [-\gamma] \cap [i\gamma]$, then
$-iz \in [\gamma] \cap [i\gamma]$.
In all cases, $[\gamma] \cap [i\gamma] \ne \emptyset$.
