Countable number of arcs between two intervals

I will argue that the answer is no: there may be no such $A$.

Lemma: there is a countable set of disjoint arcs from $I_0$ to $I_1$ such that any arc from $I_0$ to $I_1$ disjoint from this set has only finitely many $1$'s and $2$'s in the base-4 representation of its startpoint in $I_0$

I will index the arcs using rational numbers $q=a/4^b\in[0,1]$, where $A_q$ will go from $(0,q)$ to $(1,q).$ The arcs will be constructed using induction on $b$ (restricting the numerator $a$ to not be a multiple of $4$, except for $q=0=0/4^0$).

For the base case $b=0$, take a straight line for the arc from $(0,0)$ to $(1,0)$, and a straight line for the arc from $(0,1)$ to $(1,1)$.

Now assume $A_{a/4^b}$ and $A_{(a+1)/4^b}$ are given; these bound an open disc and we are free to choose sufficiently wiggly $A_{(4a+1)/4^{b+1}}$, $A_{(4a+2)/4^{b+1}}$, and $A_{(4a+3)/4^{b+1}}$ in this region to guarantee that the $x$-coordinate of any arc trapped between $A_{(4a+1)/4^{b+1}}$ and $A_{(4a+3)/4^{b+1}}$ switches between $1/4$ and $3/4$ at least $b$ times.

Consider a new arc $A$ with starting point $(0,r)$, disjoint from all the $A_q$ constructed above. By construction, for each $b$, if the $b$'th base-4 digit of $r$ is $1$ or $2$, then $A$ switches between $1/4$ and $3/4$ at least $b$ times. So $r$ cannot have infinitely many $1$'s or $2$'s in its base-4 representation, QED.

Full result

Plonk a scaled copy of the above construction into $[0,1/3]\times [0,1]$, and another copy in $[2/3,1]\times[0,1]$, restricting the second copy to arcs with $q\leq 1/2$. Join the first copy's $A_q$ to the second copy's $A_{q/2}$ by a straight line from $(1/3,q)$ to $(2/3,q/2)$. This rules out all the remaining starting points because the only numbers $q$ such that $q$ and $q/2$ both have only finitely many $1$'s and $2$'s in the base-4 representations are of the form $a/4^b$.