TITLE: Countable number of arcs between two intervals
QUESTION [7 upvotes]: First consider the two intervals $I_0=\{0\}\times [0,1]$ and $I_1=\{1\}\times [0,1]$ in the plane.  
Suppose that, for each $n\in\omega=\{0,1,2,...\}$, $A_n$ is an arc (a homeomorphic image of $[0,1]$) contained in $\mathbb [0,1]^2 \setminus (I_0\cup I_1)$ except for having one endpoint in $I_0$ and the other endpoint in $I_1$. 
And suppose $A_i\cap A_j=\varnothing$ when $i\neq j$.
We may assume $A_0=[0,1]\times \{0\}$ and $A_1=[0,1]\times \{1\}$.
It seems intuitively true that there should be an arc $A\subseteq [0,1]^2$ between $I_0$ and $I_1$ that misses $\bigcup _{n\in\omega} A_n$.
Is this true?

MY ATTEMPT: 
First consider the entire closed connected region $T_0=[0,1]^2$
  between $A_0$ and $A_1$ (as defined above). 
If there is no other $A_n$ in $T_0$, then we're done (the arc $A$ is
  easy to find). 
Otherwise, choose the least $n>1$ such that $A_n\subseteq T_0$.
Get a tube $T_1$ (homeomorphic to $[0,1]^2$) between $A_1$ and $A_n$.
  So $T_1\cap A_0=\varnothing$.  
Now get $T_2$ by going from $A_n$ to $A_m$, where $m>n$ is least index
  of any arc $A_m\subseteq T_1$.
We can continue this process to get a nested sequence of compact
  connected tubes $T_n$, so that $T_n$ misses $A_{n-1}$ for each
  $n<\omega$.  
Then $T:=\bigcap T_n$ is a compact connected set from $I_0$ to $I_1$
  that misses all $A_n$'s.   But does it contain an arc from $I_0$ to $I_1$?

EDIT (June 23): There are now two answers below, which claim opposite things.  I'll try to figure out which argument is flawed, but would appreciate any help.  Please note that I proved (above) that there is a continuum between $I_0$ and $I_1$ which misses all arcs $A_n$.  So does this mean @Colin 's answer is flawed?
Regarding @Santana 's answer:

Think about a sequence of arcs doing this. So $$A_0=[0,1]\times \{0\}$$ and for $n>0$, $$A_n=\text{graph of the function } f_n:[0,1]\to \mathbb R \text{ where }f_n (x)=\frac{1}{2}|2x-1|+\frac{1}{2^n}.$$ 
This is not homeomorphic to any collection of horizontal arcs. Moreover this shows you cannot start at an arbitrary remaining endpoint.  Every arc beginning at a point in $\{0\}\times (0,1/2)$ and ending in $I_1$ must intersect some $A_n$.

REPLY [3 votes]: 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 $a/4^b$.