$\newcommand{\R}{\mathbb R}$Here is an elementary (albeit somewhat longish) solution, without using the Jordan curve theorem.

Let us borrow the inversion idea from Olivier Bégassat, so that the two arcs be in $D$, with endpoints $-1,1$ and $-i,i$. Inscribing the disk into the closed square with horizontal and vertical sides, we see that it is enough to prove the following:

**Claim 1:** Let $R$ be a closed rectangle with horizontal and vertical sides; in the sequel, we will only consider such rectangles. Let $A$ be an arc in $R$ connecting the two vertical sides of $R$. Let $B$ be an arc in $R$ connecting the two horizontal sides of $R$. Then $A$ and $B$ intersect.

*Proof:* We have $A=a([0,1])$ and $B=b([0,1])$ for some continuous functions $a$ and $b$ from $[0,1]$ to $\R^2$ such that $a(0)\in\{0\}\times[0,1]$, $a(1)\in\{1\}\times[0,1]$, $b(0)\in[0,1]\times\{0\}$, $b(1)\in[0,1]\times\{1\}$.

Let us say that a sub-rectangle $Q=[x_1,x_2]\times[y_1,y_2]$ of $R$ is good if for some subintervals $[s_1,s_2]$ and $[t_1,t_2]$ of $[0,1]$ we have $a([s_1,s_2])\subseteq Q$, $b([t_1,t_2])\subseteq Q$, $a(s_1)\in\{x_1\}\times[y_1,y_2]$, $a(s_2)\in\{x_2\}\times[y_1,y_2]$, $a(s_1)\in\{x_1\}\times[y_1,y_2]$, $a(s_2)\in\{x_2\}\times[y_1,y_2]$. That is, the goodness of $Q$ is witnessed by the subarc $a([s_1,s_2])\subseteq Q$ (of the arc $A=a([0,1])$) connecting the left and right sides of $Q$, and by the subarc $b([t_1,t_2])\subseteq Q$ (of the arc $B=b([0,1])$) connecting the bottom and top sides of $Q$.

By compactness and continuity, there is a good sub-rectangle $Q$ of $R$ with the smallest perimeter; let us call such a good sub-rectangle $Q$ minimal.

Without loss of generality (wlog) $R$ is a minimal good sub-rectangle of itself. The case when the area of $R$ is $0$ is easy. So, wlog the area of $R$ is $>0$.

Moreover, by approximation and compactness, the connecting arcs $A$ and $B$ in $R$ are wlog polygonal (chains), say $A_0\cdots A_m$ and $B_0\cdots B_n$ for some natural $m$ and $n$ and some points $A_0,\ldots,A_m,B_0,\ldots,B_n$ in $R$ such that the points $A_0,A_m,B_0,B_n$ are respectively on the left, right, bottom, and top sides of $R$.

Let us prove the so-reduced Claim 1 (with a minimal $R$ and polygonal $A$ and $B$) by induction on $m+n$. The smallest value of $m+n$ is $2$, and then $m=n=1$, in which case Claim 1 is a simple exercise.

If for some $j\in\{1,\dots,m-1\}$ the point $A_j$ is on the left side of $R$, then we can replace the arc $A_0\cdots A_m$ by $A_j\cdots A_m$ and then use the induction on $m+n$. So, wlog the points $A_1,\dots,A_{m-1}$ are not on the left side of $R$; similarly, these points are not on the right side of $R$. Similarly, the points $B_1,\dots,B_{n-1}$ are not on the bottom or top side of $R$.

So, we can move the left side of $R$ slightly to the right and thereby get a contradiction with the minimality of $R$ unless a point $B_l$ is on the left side of $R$ for some $l\in\{0,\dots,n\}$. So, we do have a point $B_l$ is on the left side of $R$ for some $l\in\{0,\dots,n\}$. Similarly, we have a point $B_r$ is on the right side of $R$ for some $r\in\{0,\dots,n\}$. Similarly, we have a point $A_b$ on the bottom side of $R$ for some $b\in\{0,\dots,m\}$ and a point $A_t$ on the top side of $R$ for some $t\in\{0,\dots,m\}$.

The arc $B_l\cdots B_r$ (possibly with $r<l$) connects the left and right sides of $R$, and the arc $A_b\cdots A_t$ (possibly with $t<b$) connects the bottom and top sides of $R$. So, if $l\notin\{0,n\}$, then the arcs $B_l\cdots B_r$ and $A_b\cdots A_t$ intersect by the induction on $m+n$, and we are done.

So, wlog $l\in\{0,n\}$. Similarly, wlog $r\in\{0,n\}\setminus\{l\}$, $b\in\{0,m\}$, $t\in\{0,m\}\setminus\{b\}$. Also, by shifting and rescaling, wlog the rectangle $R$ of area $>0$ is the unit square $[0,1]^2$. So, if the arcs $A$ and $B$ do not intersect, then wlog $B_0=(0,0)$, $B_n=(1,1)$, $A_0=(0,1)$, $A_m=(1,0)$. Let $k,-u,v,-w$ stand for the slope coefficients of the lines $B_0B_1,A_0A_1,B_nB_{n-1},A_m A_{m-1}$, respectively. Note that $B_{n-1}$ is in $R$ but not on the top side of $R$ and is not on the right side of $R$. So, $v\in(0,\infty)$. Similarly, $k,u,w$ are each in $(0,\infty)$.

Wlog $uw\ge kv$.
Then for all small enough real $x>0$ the rectangle $R'$ with vertices $B'_0:=(x,kx)$, $B'_n=(1-\frac uv\,x,1-ux)$, $A'_0:=(x,1-ux)$, $A''_m:=(1-\frac uv\,x,kx)$ is good -- witnessed by the subarc $A'_0A_1\cdots A_{m-1}A'_m$ with $A'_m:=(1-\frac uv\,x,w\frac uv\,x)$ (of the arc $A=A_0A_1\cdots A_{m-1}A_m$) connecting the left and right sides of $R'$ and by the subarc $B'_0B_1\cdots B_{n-1}B'_n$ (of the arc $B=B_0B_1\cdots B_{n-1}B_n$) connecting the bottom and top sides of $R'$.

Thus, we get a good rectangle $R'$ whose perimeter is smaller than that of the minimal rectangle $R$, the final contradiction. $\quad\Box$

Here is a picture showing the rectangle $R$ (which was wlog the unit square), the rectangle $R'$ (dotted), and the points $A_0,A_m,B_0,B_m,A'_0,A''_m,A'_m,B'_0,B'_m$ as in the penultimate paragraph of the proof with $(k,u,v,w,x)=(1, 2, 1, 3,1/20)$:

Messenger (2)6(1877), 132–133. JFM 09.0393.01. $\endgroup$2more comments