Why is this particle interaction forbidden? Why is this interaction forbidden?
$$\nu_e+\bar{\nu}_e\rightarrow K^+ + K^-$$
Lepton number is conserved, charge is conserved, baryon number is irrelevant since these are mesons. Energy is conserved since we could collide these two neutrinos (and anti) at any energy required to produce the two mesons. Neutrinos only interact via the weak force so if it were possible it would be a weak decay. Parity needn't be conserved since this would be a weak interaction. The only thing that is bugging me is that quark flavour (and strangeness) is not conserved but I thought this was allowed via a weak interaction. Maybe it is momentum?
EDIT 
I think this process is allowed and here is my attempt at the lowest order Feynman diagram, does this look correct?

EDIT 
Second attempt at a Feynman diagram, following comments made.

 A: Your first diagram is wrong, since there is no vertex in a Lorentz invariant theory where three fermions and a vector meet.
However, I don't see why you say the interaction is forbidden. It would surely be insanely suppressed since amplitudes are extremely weak, but I don't see a problem with the diagram (for instance):

Notice that quarks mix, so the vertex $Wsu$ is there, albeit suppressed by an off-diagonal element of the Cabibbo-Kobayashi-Maskawa quark mixing matrix. The interaction is allowed.
(I just saw the second diagram you drew, which is another valid -- and even much more relevant in magnitude -- contribution to the full amplitude).
EDIT: A nice reference for the Standard Model which IMHO may clarify beginner course confusion is Giunti & Kim's "Fundamentals of Neutrino Physics and Astrophysics" (2007, Oxford University Press), Chapter 3.  
A: I am very bad at drawingin "paint", but the process can go as 
antineutrino +neutrino to Z0 , Z0 to  s  antis (or up antiup), a gluon vertex from the quark to  an up antiup (or strange antistrange quark) the parenthesis are the alternate diagram. ,

So it is not forbidden, it has two weak vertices and so very small cross section.
A: The problem with this process is, that there is no term in the standard model which allows a coupling of a fermion a W-boson and a quark-antiquark pair.
Even with the neutrino having no charge, you have to take this arrow you draw on the fermion line seriously, so the fermion flow is violated at this interaction points. Furthermore you can only couple neutrinos with the presence of an electron or positron respectively (same for higher lepton families)
