Are all computable numbers constructable after a countable number of steps? While looking at another question on this site about constructable numbers I started wondering.  If you can take a countable number of steps (possibly infinite) can you draw an interval of a length corresponding to a computable number?
More strictly if I have a unit interval, a straight edge, a compass, a finite list of instructions (which can include instructions to repeat sequences of the instructions until an event occurs, instructions to draw lines using my tools and instructions on labeling points) and the ability to carry out a countably infinite number of actions in a finite time.  Can I construct a interval that corresponds to a given computable number?
 A: If you can take an infinite number of steps you construct a non-constructible length?
No. If you take infinitely many steps, you never finish drawing and hence you cannot draw what you want.
However, you might want to ask whether you can draw successive approximations that get arbitrarily close to the desired drawing (such as a segment with its length being a certain value relative to a given unit segment). In that case...
It depends. If you just ask for the existence of a sequence of successive approximations, then...
Yes even for non-computable numbers. Any real number has such a sequence because every real number is the limit of some sequence of rational numbers, and all rational numbers are constructible (by compass and straightedge).
But if you ask for the existence of a computable sequence of approximations then...
Yes only for computable numbers. Every computable number is by definition produced digit by digit by a procedure, and every time you get a new digit you can construct the corresponding approximation that is accurate to that digit. It can certainly be done because the approximations are all rational. Conversely, if you have a computable sequence of approximations that converge to a length, then clearly the limit is computable.
A: You can certainly define a (possibly infinite) sequence of segments whose total length has as its limit any computable number.  You can compute the binary expansion of the number.  The integer part is easy, just add up the proper number of $1$'s.  Then add on $\frac 1{2^n}$ if the $n^{\text{th}}$ bit of the expansion is $1$.  This will give you the correct limit.
