How to make a symmetrical DC power source?
Any ideas on why this circuit doesn't work?


Oh yeah. In one word (or at least one number): 741.

Never, ever, try to use a 741 with power supplies much different from +/- 15 volts. +/-12 is probably OK, and I've seen +/- 9 used. +/- 2.5? Hahahahaha.

Sorry. I got carried away.

But seriously, that's your problem (or at least the most obvious one). If you really must operate from a 5 volt supply, you need to use a comparator (not an op amp) rated for that voltage.

And your LED circuit is asking for trouble. If you use an LED with Vf less than about 2.5 volts and a rail-to-rail drive, you risk killing the LED. If you use something like a white LED, with a 3 - 3.5 volt threshold, you will get little or no output from the LED. 

Umm. And let's talk about your 5 volt supply. You realize, I hope, that you will be tempted to use the same 5 volt supply for your logic as your probe. And if you do you cannot possibly establish your virtual ground. Look at your schematic again, and put a ground connection on the - side of the 5 volts, and think about what will happen.

Finally, be aware that the LED current will affect the midpoint ground set by the two 1k resistors. In fact, if you have (let's say) 10 mA through the LED, this is 4 times the nominal current through the resistors and will swamp them.

EDIT - In comment, an alternative (LM339) was suggested. 

That won't work either, and in some respects is an even worse choice than a 741. Regardless of supply voltages, it cannot drive the High LED at all. The output of the LM339 is essentially a one-way switch between the output and the - supply (called ground on a 339). When active (LOW output) it will turn on the Low LED. When not (HIGH output) it will do nothing at all, so the High LED will never turn on. See page 10 of the data sheet.

It also has considerably higher current capacity than a 741, so is almost guaranteed to kill the Low LED if it does turn it on. LEDs need current limiting.