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Here is a sketch of the proof of item 2. [Edit, Nov 17, 2014: My original argument for item 3 started with conditionally convergent series $\sum_i u_i=r$ and, for each $s\ge r$, produced a rearrangement fixing the positive terms, and whose partial sums have liminf $s$, but to argue that their limsup is also $s$ requires additional work. Fixing this lead me to a proof very similar to Sierpiński's, so I decided to reorganize the answer and just present his argument, together with what I hope are some additional explanations. What follows is part of a post on rearrangements of series on my blog.]

Apply the lemma to the $a_i$ to obtain a rearrangement $c_0,c_1,\dots$ with partial sums $C_0,C_1,\dots$ such that $A_i-C_i\to L$. Consider the rearrangement $v_0,v_1,\dots$ of the $u_i$ that fixes the $-b_i$ and permutes the $a_i$ as indicated. If $V_0,V_1,\dots$ are the partial sums of the $V_i$, note that, by design, $V_i=C_{p_i}-B_{n_i}$, so that $U_i-V_i=A_{p_i}-C_{p_i}\to L$, and $V_i\to U-L=V$, as wanted. All that remains is to prove the lemma.

Note that our assumption that the $a_k$ are  decreasing ensures that, in case 2, (with notation as above) $A_n-C_n=(A_{n-1}-C_{n-1})+(a_n-c_n)\le$ $A_{n-1}-C_{n-1}$, since $a_i\ge a_n$. Moreover, it cannot be that we have equality from some point on, since the $a_k$ converge to $0$. This means that the values $A_k-C_k$ decrease.

Here is a sketch of the proof of item 2. [Edit, Nov 17, 2014: My original argument for item 3 started with conditionally convergent series $\sum_i u_i=r$ and, for each $s\ge r$, produced a rearrangement fixing the positive terms, and whose partial sums have $\liminf s$, but to argue that their $\limsup$ is also $s$ requires additional work. Fixing this lead me to a proof very similar to Sierpiński's, so I decided to reorganize the answer and just present his argument, together with what I hope are some additional explanations. What follows is part of a post on rearrangements of series on my blog.]

Apply the lemma to the $a_i$ to obtain a rearrangement $c_0,c_1,\dots$ with partial sums $C_0,C_1,\dots$ such that $A_i-C_i\to L$. Consider the rearrangement $v_0,v_1,\dots$ of the $u_i$ that fixes the $-b_i$ and permutes the $a_i$ as indicated. If $V_0,V_1,\dots$ are the partial sums of the $v_i$, note that, by design, $V_i=C_{p_i}-B_{n_i}$, so that $U_i-V_i=A_{p_i}-C_{p_i}\to L$, and $V_i\to U-L=V$, as wanted. All that remains is to prove the lemma.

Note that our assumption that the $a_k$ are decreasing ensures that, in case 2, (with notation as above) $A_n-C_n=(A_{n-1}-C_{n-1})+(a_n-c_n)\le$ $A_{n-1}-C_{n-1}$, since $a_i\ge a_n$. Moreover, it cannot be that we have equality from some point on, since the $a_k$ converge to $0$. This means that the values $A_k-C_k$ decrease.

Here is a sketch of the proof of item 2. [Edit, Nov 17, 2014: My original argument for item 3 started with conditionally convergent series $\sum_i u_i=r$ and, for each $s\ge r$, produced a rearrangement fixing the positive terms, and whose partial sums have liminf $s$, but to argue that their liminf is also $s$ requires additional work. Fixing this lead me to a proof very similar to Sierpiński's, so I decided to reorganize the answer and just present his argument, together with what I hope are some additional explanations. What follows is part of a post on rearrangements of series on my blog.]

Here is a sketch of the proof of item 2. [Edit, Nov 17, 2014: My original argument for item 3 started with conditionally convergent series $\sum_i u_i=r$ and, for each $s\ge r$, produced a rearrangement fixing the positive terms, and whose partial sums have liminf $s$, but to argue that their limsup is also $s$ requires additional work. Fixing this lead me to a proof very similar to Sierpiński's, so I decided to reorganize the answer and just present his argument, together with what I hope are some additional explanations. What follows is part of a post on rearrangements of series on my blog.]

Since $a_i,c_i\to 0$, it follows that, if case 1 is also applied infinitely often, then $A_i-C_i\to L$.  To conclude, it suffices therefore to argue that we cannot stay in case 2 forever. To see this, argue by contradiction, and assume that from $n-1$ on, we always stay in case 2. Let $j_m$ be the number of indices $i\le m$ such that $a_i$ is not one of the $c_k$, $k\le m$. For any $m\ge n$, since $A_{m-1}-C_{m-1}>L$, then $j_m\le j_{m-1}$, with equality if and only if $a_m$ is not one of the $c_k$, $k<m$. Since $C_{n-1}\ne A_{n-1}$, necessarily at least one of the $c_i$, $i\le n-1$, is $a_r$ for some $r\ge n$. If $r_0$ is the least such index $r$, then $j_{r_0}<j_{n-1}$, because $a_r$ is one of the $c_i$ with $i<r$, in fact $i<n$. Since we cannot have an infinite decreasing sequence of positive integers, this means that (we have a contradiction and) eventually we should be in case 1 again. This completes the proof in the case the $a_i$ are decreasing.

1. If $A_{n-1}-C_{n-1}\le L$, as before let $c_n$ be some $a_r$ where the index $r$ has not yet been chosen, and $a_r<a_n/2$, and $a_r<1/2^n$.
2. If $A_{n-1}-C_{n-1}>L$, now we consider two possibilities, according to whether or not the index $n$ was picked previously: If it was not, then we let $c_n=a_n$. Note that $A_n-C_n=A_{n-1}-C_{n-1}$ if this is the case, and that, since $C_{n-1}\ne A_{n-1}$, then necessarily some $c_i$ with $i<n$ must be $a_r$ for some $r>n$. This means that at some later stage (at most by stage $r$, we are no longer to be in this case.
3. If $A_{n-1}-C_{n-1}>L$, and the index $n$ was chosen previously, then we let $c_n$ be $a_i$, where $i$ is the first index less than $n$ not chosen yet.

If stage $n$ is by case 1, and stage $m<n$ was largest where $A_m-C_m$ ${}>L$, then $A_{m+1}-C_{m+1}=(A_{m+1}-C_{m+1})+(a_{m+1}-c_{m+1})>$ $L-c_{m+1}>L-\epsilon$. Since the sequence $A_i-C_i$ is increasing for $m+1\le i\le n$, we have $L\ge A_n-C_n>L-\epsilon$, as wanted.

Suppose then that stage $n$ is by case 3, and that stage $m<n$ was largest where we were in case 1. We then have that $A_{m+1}-C_{m+1}=(A_m-C_m)+(a_m-c_m)\le L+a_m<L+\epsilon$. Also,

$\displaystyle \sum_{i=m+2}^n(a_i-c_i)=\sum\{a_i-c_i\mid m+2\le i\le n,$ and stage $i$ was by case 3$\}$ $\displaystyle \le\sum\{a_i\mid m+2\le i\le n,$ and stage $i$ was by case 3$\}$ $\displaystyle \le\sum_{j>N}\frac1{2^j}<\epsilon.$

This means that $A_n-C_n<L+2\epsilon$. But also $A_n-C_n=(A_{n-1}-C_{n-1})+(a_n-c_n)>L-c_n>L-\epsilon$. This completes the proof. $\Box$

Since $a_i,c_i\to 0$, it follows that, if case 1 is also applied infinitely often, then $A_i-C_i\to L$.  Therefore, to conclude, it suffices to argue that we cannot stay in case 2 forever. To see this, argue by contradiction, and assume that from $n-1$ on, we always stay in case 2. Let $j_m$ be the number of indices $i\le m$ such that $a_i$ is not one of the $c_k$, $k\le m$. For any $m\ge n$, since $A_{m-1}-C_{m-1}>L$, then $j_m\le j_{m-1}$, with equality if and only if $a_m$ is not one of the $c_k$, $k<m$. Since $C_{n-1}\ne A_{n-1}$, necessarily at least one of the $c_i$, $i\le n-1$, is $a_r$ for some $r\ge n$. If $r_0$ is the least such index $r$, then $j_{r_0}<j_{n-1}$, because $a_{r_0}$ is one of the $c_i$ with $i<r_0$, in fact $i<n$. Since we cannot have an infinite decreasing sequence of positive integers, this means that (we have a contradiction and) eventually we should be in case 1 again. This completes the proof in the case the $a_i$ are decreasing.

1. If $A_{n-1}-C_{n-1}\le L$, as before let $c_n$ be some $a_r$ where the index $r$ has not yet been chosen, and $a_r<a_n/2$, and $a_r<1/2^n$.
2. If $A_{n-1}-C_{n-1}>L$, now we consider two possibilities, according to whether or not the index $n$ was picked previously: If it was not, then we let $c_n=a_n$. Note that $A_n-C_n=A_{n-1}-C_{n-1}$ if this is the case, and that, since $C_{n-1}\ne A_{n-1}$, then necessarily some $c_i$ with $i<n$ must be $a_r$ for some $r>n$. This means that at some later stage (at most by stage $r$), we are no longer to be in this case.
3. If $A_{n-1}-C_{n-1}>L$, and the index $n$ was chosen previously, then we let $c_n$ be $a_i$, where $i$ is the first index less than $n$ not chosen yet.

If stage $n$ is by case 1, and stage $m<n$ was largest where $A_m-C_m$ ${}>L$, then $A_{m+1}-C_{m+1}=(A_m-C_m)+(a_{m+1}-c_{m+1})>$ $L-c_{m+1}>L-\epsilon$. Since the sequence $A_i-C_i$ is increasing for $m+1\le i\le n$, we have $L\ge A_n-C_n>L-\epsilon$, as wanted.

Suppose then that stage $n$ is by case 3, and that stage $m<n$ was largest where we were in case 1. We then have that $A_{m+1}-C_{m+1}=(A_m-C_m)+(a_m-c_m)\le L+a_m<L+\epsilon$. Also, $$A_n-C_n=A_{m+1}-C_{m+1}+\sum_{i=m+2}^n(a_i-c_i),$$ but for any such $i$, we have that stage $i$ was either by case 2, and therefore $a_i-c_i=0$, or else it was by case 3, and therefore $a_i=c_j<1/2^j$ for some $j<i$, but necessarily $j>N$, so $$\sum_{i=m+2}^n(a_i-c_i)=\sum\{a_i-c_i\mid m+2\le i\le n,\mbox{ and stage }i\mbox{ was by case 3}\}\le\sum\{a_i\mid m+2\le i\le n,\mbox{ and stage }i\mbox{ was by case 3}\} \le\sum_{j>N}\frac1{2^j}<\epsilon.$$

This means that $A_n-C_n<L+2\epsilon$. But also $$A_n-C_n=(A_{n-1}-C_{n-1})+(a_n-c_n)>L-c_n>L-\epsilon.$$ This completes the proof. $\Box$