Is there always a continuous extension of a countable collection of quotient maps, whose domains are a Haussdorff-convergent sequence? Let $(\mathbf{B},\|\cdot\|)$ be a Banach space, $(A_{n})$,$(B_{n})$ collections of compact, connected subsets of $\mathbf{B}$ and let $\{q_{n}:A_{n}\to B_{n} : n\geq 1\}$ be a collection of quotient maps (closed, continuous, surjective).
The Hausdorff distance $d_{H}$ is a metric on the collection of closed and bounded subsets of $\mathbf{B}$.
$$
d_{H}(A,B)= \max\left\{   \sup_{b\in B} \|b-A\|   \,,\,   \sup_{a\in A} \|a-B\|   \right\}
$$
where $\|b-A\|=\inf\{\|b-a\| : a\in A\}$ as usual.
If $d_{H}(A_{n},A)\to 0$ and $A_{1},A_{2},A_{3}, \ldots, A$ are mutually disjoint:
(Q1) Does this implies the existence of a continuous function 
$$ P:\bigsqcup A_{n} \to \bigsqcup B_{n} $$ 
such that $P_{|A_{n}} = q_{n}$ for every $n\geq 1$?
(Q2) Suppose $P$ exists. Is the Hausdorff limit $A$ contained in the domain, 
   or does $P$ should be extended further?  
(Q3) Suppose $P$ had to be extended in its domain to contain $A$,what can we say about $P_{|A}$?
(Q4) If the quotient maps $q_{n}:A_{n} \to B_{n}$ are Lipschitz continuous (or some other type of strong continuity), does that makes any difference?
$\mathbf{New \, questions}$ (20 Jun):  Let $A'=\bigsqcup A_{n}$
I would like to characterize Hausdorff convergence of the sequence ${P(A_{n})}={B_{n}}$.  
When $P$ can be extended $\mathtt{continuously}$ to $(A' \sqcup A)$ its known that $d_{H}(A_{n},A) \to 0$ implies $d_{H}(P(A_{n}),P(A))= d_{H}(B_{n},P(A))\to 0$. This continuous extension exists whenever $\{x_{n}\}\subset A'$ converges in $A' \sqcup A$ then $\{P(x_{n})\}$ converges. 
Suppose $P$ has not yet been extended to contain $A$ in its domain, and let $q:A\to B$ be a quotient map for some $B\subset \mathbf{B}$, then:
(Q5) Since $A$ and $A'$ are disjoint and define a locally finite family: There a continuous extension of $P$ and $q$? What is happening here?
(Q6) If all the quotient are Lipschitz, i.e. $\|q_{n}(x)\| \leq \|x\|$ for every $x\in A_{n}$. Does convergence of a sequence $\{x_{n}\}$, and therefore boundedness of $\{P(x_{n})\}$, implies some sort of Hausdorff convergence for the sequence $P(A_{n})=B_{n}$?
 A: Q1. Yes. Put $A’=\bigsqcup A_{n}$. The family $\{A_n\}$ is locally finite in $A’$. Indeed, let $x\in A’$ be an arbitrary point. Since $x\not\in A$ and the set $A$ is closed, $\varepsilon=\|x-A\|/2>0$. Since the sequence $\{A_n\}$ converges to $A$, there exists a number $N$ such that $d_H(A_n, A)<\varepsilon$ for each $n>N$. Thus the $\varepsilon$-neighborhood of the point $x$ if disjoint from the set $A_n$ for each $n>N$. Now the continuity of the map $P$ follows from [Eng, Prop. 2.1.13] 



Q2. Yes, of course, because by definition domain $\bigsqcup A_{n}$ of $P$ is disjoint from $A$. But extension of $P$ to $A$ exists iff for every sequence  $\{x_n\}$ of points of $A’$convergent in  $A’\cup A$ ,  the sequence $\{ P(x_n )\}$ is also convergent (see, for instance, [MR, Lemma 2]. It not always exists. For instance, let $(\mathbf{B},\|\cdot\|)$ be the space $\Bbb R$ endowed with the standard norm, $A_n=\{1/n\}$, $B_n=\{n\}$, and $A=\{0\}$. 
Q3. Can you tell more precisely in which properties of $P$ you are interested?
Q4. I guess, no.
References
[MR] M. A. Mytrofanov, A. V. Ravsky, Approximation of continuous functions on Fréchet spaces, Journal of Mathematical Sciences, 185:6 (2012), 792--799
[Eng]  Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
