Spivak's "Differential Geometry" Volume 1, Chapter 1 ,Problem #20 part (b)

For those not looking at the diagram, the "infinite jail cell" is the boundary of a tubular neighborhood of the integer grid (including horizontal a vertical lines) in the $z = 0$ plane of $xyz$-space.

If we take a tubular neighborhood of the line segments from $(0,0)$ to $(1, 0)$ to $(1,1)$ to $(0, 1)$ to $(0,0)$, we get something toroidal. The "cylinder" described by Spivak consists of half of each meridian of this torus -- the half that's closest to the point $(1/2, 1/2)$. Let's call that cylinder $C$; its boundary consists of two squared-off circles in the $z = \pm 1/2$ planes.

I believe that what Spivak is suggesting is that you next consider the part of the infinite jail cell that's between two larger squared-off circles, essentially ones that are parallel to the $z=0$-plane squares with corners $(-1, -1), (2, -1), (2, 2), (-1, 2)$, offset by $1/2$ in the positive and negative $z$ directions. Call this larger part $D$. It basically consists of 9 cells, the center one of which is $C$. The boundary of $D - C$ consists of four circles; the set $D-C$ is clearly a 2-manifold-with-boundary, hence homeomorphic to a $k$-holed torus with four disks removed for some $k$. (I'm pretty sure that $k$ is $8$, but I could easily be off by one or two.)

Now consider a $k+2$-holed torus, with the $k+2$ holes arranged in a line, like the infinite torus drawn by Spivak, and cut off the left and right half-tori. You end up with a $k$-holed-surface-with-boundary, having four circle-boundaries. This is clearly homeomorphic to D - C. And a single cylinder is homeomorphic to $C$, so the union, along common boundaries, gets you something that's homeomorphic to $D$.

You now proceed by induction: drawing a 5 x 5 pair of squared-off-circles, you see that the region $E$ between them and the 3x3 object $D$ is also homeomorphic to a $p$-holed chain with four circular ends, etc.

In other words, the region between adjacent circle-pairs is always homeomorphic to an $s$-holed-torus-with-boundary, having four boundary components. By the classification theorem for surface, this is the same as a truncated piece of a linear chain of tori. You thus build up a sequence of homeomorphisms

$h_i : U_i \rightarrow V_i$

where $U_i$ is a chain of $(2i + 1)^2 + 1$ tori with the last half-torus removed, and $V_i$ is the part of the infinite jail-cell enclosed by the two squared-off circles of edge-length $i$. The nice thing about this sequence is that $h_{i+1} = h_i$ on $U_i$, so the limit is well-defined and is a homeomorphism as well.

====

An alternative:

Let's just talk, from here on, about squares on the integer grid, OK? So what I did in the first solution was start with a 1x1 square, move to a concentric 3x3 square, and so on.

As an alternative, you can take, as your second square, a 2x2 square whose lower-left corner is the origin. The difference $D - C$ in this case clearly is homeomorphic to a 3-holed torus with some boundary, and the same sort of union argument applies. Now you take, as your third square, a 3x3 one whose lower-left is at $(-1, -1)$. Again, the difference $E - D$ is clearly homeomorphic to a 5-holed torus with some boundary. And in general, you can extend like this, alternating between adding a band to the northeast and adding a band to the southwest, gradually handling all the cells of the infinite jail-window.

The advantage of the second approach is the obvious homeomorphism between the added strip and a $2i+1$-holed torus with some disks removed. The disadvantage is that the gluing-map is more complex. I think it's about a wash.