Separating boric acid from sodium chloride I want to create some boric acid with borax in the following reaction:

$$\ce{Na2B4O7·10H2O + 2 HCl → 4 BH3O3 + 2 NaCl + 5 H2O}$$ 

However as you can see, the leftover solution will contain a mixture of boric acid, sodium chloride and water. Both chemicals are soluble in cold water so neither will want to immediately crash out of solution.

My question is, how can I remove the excess water and sodium chloride to leave myself with some powdered/dry boric acid crystals?

I know I can probably just use some liquid ammonia, but then the boric acid is only slightly soluble in it leaving much of it behind. I could perhaps use glycerin also, however I am limited to how much I currently have.
Here is a potential solution that shouldn't even require heating. Note that I have not tried this myself. 

Boric acid is soluble in lower alcohols like ethyl alcohol and isopropyl alcohol.
The solubility of sodium chloride in ethyl alcohol at room temperature is about 0.05%$^1$.

Given these properties, I suggest dissolving the dried boric acid / sodium chloride in a minimal amount of cold 95% denatured ethyl alcohol or 99% isopropyl alcohol (available at most pharmacies).  

The boric acid should dissolve readily and the sodium chloride should not, so you can just filter it and let it dry to a fairly sodium chloride free boric acid solid.  

To make the sodium chloride even less soluble you could cool the solution in a freezer prior to filtering.  If you notice any significant increase in solids that means boric acid is coming out of solution and you'll want to abort this step.

1) Journal of Chemical and Engineering Data, Vol. 50, No. 1, 2005