Question: Consider an electron emitting a single photon. There is exactly one gamma matrix (corresponding to the photon vertex) between the outgoing spinor $\bar{u}$ and incoming spinor $u$.
This implies that both $\bar{u}$ and $u$ have to be right handed electrons (or both left handed). 

In other words the non-zero amplitude corresponds to 
$u^{\dagger}(\frac{1 + \gamma^5}{2}) \gamma^0 \gamma^{\mu} (\frac{1 + \gamma^5}{2}) u$. 

The amplitude of $u^{\dagger}(\frac{1 - \gamma^5}{2}) \gamma^0 \gamma^{\mu} (\frac{1 + \gamma^5}{2}) u$ would be 0.

Lets ignore the mass of the electron giving the helicities to be frame independent.

This means that both outgoing and incoming electrons have to have either spin +1/2 or spin -1/2.

The question is: when the +1/2 electron emits a photon (of spin +1) then, should not the outgoing electron have spin -1/2; so that spin is conserved?
What/where is the flaw in my analysis?

Answer: 
I think that although spin conservation is a good rule of the thumb, it is not necessary. What should be conserved at all times is the total angular momentum: J=S+L (S:spin,L=angular momentum). Therefore after the emission of the photon J should be the same as before and not S...
An electron running along does not radiate. A photon can be emitted if there is an interaction ( what acceleration means for classical electrodynamics), i.e.energy has to be supplied which will be released as radiation. An example is [brehmsstrahlung radiation](http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/xrayc.html), 

[![ brems](https://i.stack.imgur.com/kBXX5.jpg)](https://i.stack.imgur.com/kBXX5.jpg)

The virtual photon in the diagram can be exchanged with a magnetic or electric field in a more general case of acceleration and radiation.

So it is not a simple conservation of angular momentum or an electron and an emitted photon, as more elements have to enter the interaction introducing angular momenta, and it is angular momentum that has to be conserved , as @hal states.