A question with infinity
There are various ways to do this. Here's one: $$ \begin{array}{rrrrrrrrrrrrrrrr} 1 & \rightarrow & 2 & & 6 & \rightarrow & 7 & & 15 & \rightarrow & 16 \\ & \swarrow & & \nearrow & & \swarrow & & \nearrow & & \swarrow \\ 3 & & 5 & & 8 & & 14 \\ \downarrow & \nearrow & & \swarrow & & \nearrow \\ 4 & & 9 & & 13 \\ & \swarrow & & \nearrow \\ 10 & & 12 \\ \downarrow & \nearrow \\ 11 \end{array} $$
Here's another: $$ \begin{array}{rrrrrrrrrrrrrrr} 1 & \rightarrow & 2 & & 9 & \rightarrow & 10 & & 25 & \rightarrow \\ & & \downarrow & & \uparrow & & \downarrow & & \uparrow \\ 4 & \leftarrow & 3 & & 8 & & 11 & & 24 \\ \downarrow & & & & \uparrow & & \downarrow & & \uparrow \\ 5 & \rightarrow & 6 & \rightarrow & 7 & & 12 & & 23 \\ & & & & & & \downarrow & & \uparrow \\ 16 & \leftarrow & 15 & \leftarrow & 14 & \leftarrow & 13 & & 22 \\ \downarrow & & & & & & & & \uparrow \\ 17 & \rightarrow & 18 & \rightarrow & 19 & \rightarrow & 20 & \rightarrow & 21 \end{array} $$
A simpler formula to problem 3 is to move row $r$, column $c$ to box $2^{r-1}(2c-1)$. For three dimensions you can move row $r$, column $c$, layer $l$ to box $2^{l-1}(2^{r}(2c-1)-1)$. Do you see why these work and how to generalize to $n$ dimensions?
Let me comment on your solution and some of those proposed in the other answers. First of all, both of Michael Hardy's pictures will lead to formulas more complicated than yours, because he goes back and forth rather than always going in the same direction as you did. In particular, his first picture is just the back-and-forth version of yours except that you tabulated numbers instead of drawing a picture. (For related non-mathematical amusement,you might look up "boustrophedron".) Hagen von Eitzen, on the other hand, has a really simple formula, and it actually admits a rather nice description in terms of moving people from infinitely many rows of boxes to a single row; I'll call that single row the "target" row. I haven't checked whether the following gives exactly Hagen's formula, but it'll be close. Tell the people in the first row of boxes to occupy every second box in the single target row, just as you did when you stated with two rows of boxes. They occupy target boxes $0,2,4,\dots$, leaving infinitely many empty boxes in the target row, namely $1,3,5,\dots$. Now tell the people in the second of your original rows to occupy every second one of those still-empty boxes, namely $1,5,9,\dots$, again leaving infinitely many boxes empty, namely $3,7,11,\dots$. Continue in the same way with each successive row, putting its occupants into every second one of whatever target boxes remain empty; that leaves infinitely many target boxes empty, so you can continue to the next row.
The following answer is not yet complete solution, but I think it might be an idea how to approach the problem yourself.
I recall the Collatz-problem in the "syracuse"-statement and look at the inverse transformation which allows to construct an infinite tree with infinitely many infinitely long "boxes" where each odd natural number occurs exactly one time (well, whether all odd natural numbers occur at least once is really open, but it would not spoil the resulting infinite indexing-system).
The "syracuse"-statement of the Collatz-transformation is with an odd a and b and a natural $A \ge 1$ is $$ b = {3a+1 \over 2^A} $$ where the value of A and that of b are of course uniquely determined by the value of a. But there are a lot (actually infinitely many) of different a which lead to the same b, so if we invert the transformation,
$$ b = {a 2^A - 1 \over 3} $$ we have for each a by infinitely many A's infinitely many different b, so we could write infinitely many indexed b as $$ b_{A} = {a 2^A - 1 \over 3} $$ and now each of the infinitely many $b_A$ can be treated similarly to get $$c_{A,B} = {b_A 2^B - 1 \over 3} $$
This scheme implements an infinite long index (the infinite number of boxes) and each box is of infinite length. One need now to show, that the occuring values are unique, which I think is not too difficult (it is not solving Collatz, because that (unsolved) conjecture is that the values are not only unique but also including all odd positive natural numbers which makes it so difficult).
(I have also another statement from the same problem for the infinitude of infinite bins, maybe I can reproduce it here later as an even easier-to-see representation, I'm short of time at the moment, sorry)