Is N the cts image of the Sorgenfrey line? I have this question:
Prove/disprove: The set of natural numbers (including zero) with usual topology is the continuous image of the Sorgenfrey line.
Can't we take the map $g: \mathbb{R}_{l} \rightarrow \mathbb{N}$ given by $g(x) = |[x]|$ ? (i.e the absolute value of the floor function).
 A: The OP seems to have a pretty good handle on the question and maybe just needs to think it through to be fully satisfied.
But here is a slightly different approach which makes it even more clear (to me) that the OP's method will work.  This was borne out of the remark I made (to myself) upon reading the OP's clarification that the natural numbers include zero (you're damn right they do, by the way): namely, it certainly doesn't matter, because any two countably infinite discrete spaces are homeomorphic.
Taking that one step further, it is clearly enough to realize $\mathbb{Z}$ with the discrete topology as a continuous image of the Sorgenfrey line: having done this, compose with any homeomorphism (i.e., bijection!) from $\mathbb{Z}$ to $\mathbb{N}$.  (Or, in fact, with any surjection, as the OP has done.)  For this, literally take the greatest integer function.  The preimage on any given basis element -- i.e., a singleton set $\{n\}$ -- is the half-open interval $[n,n+1)$.  I don't keep too much information about the Sorgenfrey line in my head, but I'm pretty sure those sets are open!  
A: Yes we can.  Is your question how to show that it is continuous?  Is the inverse image of $\{k\}$ open in $\mathbb{R}_l$?
