Why are all mesons unstable? My question is: why are all mesons unstable?
Shouldn't the lightest mesons like the pion be stable because they are made from the lightest quarks (and anti-quarks)? Or could the they annihilate (the quark and anti-quark)? 
 A: You are almost right about quarks and antiquarks annihilating within the meson, but for example, a π+ contains an up and an anti-down quark, and those are not antiparticles of each other.
The picture that particle physics draws, has a bunch of conserved quantities, and more or less anything that obeys all of the conservation laws happens at some rate or another. Some of these conservation laws are now known to be lies (weirdnesses to do with the Higgs mechanism and the flavor oscillations of the neutrinos and such), but I am not actually a particle physicist and so those things go above my knowledge. But I can tell you my understanding of what we know.
The intrinsic conserved quantities which each particle has are: 


*

*a baryon number B

*three lepton numbers: an electron number Ne, a muon number Nμ, and a tau number Nτ

*an isospin T (more formally the “third component of weak isospin”)

*a hypercharge Y (more formally the “weak hypercharge”)


Other conserved quantities like electric charge are a linear combination of the above. For the known matter particles (“fermions” as opposed to “bosons” which are force particles) these numbers are,
                      B   Ne   Nμ   Nτ      T      Y
e-neutrino            0    1    0    0     1/2    -1
μ-neutrino            0    0    1    0     1/2    -1
τ-neutrino            0    0    0    1     1/2    -1
electron              0    1    0    0    -1/2    -1
muon                  0    0    1    0    -1/2    -1
tau                   0    0    0    1    -1/2    -1
up/charm/top         1/3   0    0    0     1/2   +1/3
down/strange/bottom  1/3   0    0    0    -1/2   +1/3

And of course there is an antiparticle for each of these which has all of its quantum numbers above reversed in sign. So an anti-down has baryon number $-1/3$, zero for its lepton numbers, an isospin of $+1/2$ and a hypercharge of $-1/3$.
From this table, mesons have net baryon number 0, and net hypercharge 0. But they can have a net isospin of -1 (e.g. up + anti-down) or 0 (e.g. up + anti-up) or +1 (e.g. strange + anti-up.) 
Now, all matter interactions involve exactly 3 particles: two matter particles and one force particle. The force particles which two particles can annihilate into are,
     B   Ne   Nμ   Nτ    T    Y
W+   0    0    0    0   +1    0
W-   0    0    0    0   -1    0
Z0   0    0    0    0    0    0
γ    0    0    0    0   0    0

The W and Z bosons have huge masses, way larger than the pions, so the pions cannot strictly speaking decay into them. however, through the magic of quantum tunneling, the universe can use this classically forbidden state as an intermediary state to jump to a lower permanent energy level, at the cost of slowing down the process. 
The photon (represented by the Greek letter γ above) by contrast is massless and you can see direct annihilation of neutral mesons into it. However because it is massless, a single photon cannot simultaneously have energy and zero momentum: so if the quarks annihilate from rest then you actually need two photons going in opposite directions to conserve both energy and momentum. (There is another slight difference between the Z-boson and the gamma, which is that the gamma does not couple different families of particles together above, and only interacts with the charged particles, not the neutrinos.)
The π+ pion decays, similarly, but only because  its up and anti-down quark merge into a W+ boson for zero time, after which the W+ must decay into two particles, totally less massive than the π+, which also have the above sum of quantum numbers. In this case there are only two such options: positron (anti-electron) plus e-neutrino, or anti-muon plus μ-neutrino.
It turns out that the muon case is overwhelmingly more likely, and this has a complicated reason behind it which ultimately boils down to "the muon is more massive, and more massive things are easier for the W bosons to decay into." (The effect is technically called "helicity suppression".) But muons decay into electrons eventually so you eventually end up with a free π+ becoming a positron, emitting the excess mass as a bunch of neutrinos. It decays into an anti-muon first after a mean lifetime of 26 nanoseconds, then the anti-muon decays into a positron after a longer mean lifetime of about 2 microseconds, which—you guessed it—is long because it needs to decay into a μ-neutrino via a second $W^+$, and that $W^+$ then instantly decays into the positron and an e-neutrino. 
The π0 pion with 0 isospin is even more unlucky, because there is a lower-mass particle which it can decay into directly (having B = N_ = T = Y = 0), and that's the photon. You need two photons to properly conserve momentum, but that's the only subtlety. So they have an incredibly short lifespan (84 attoseconds!) because it is indeed a sort of "quark-antiquark annihilation".
But I hope that this whole comment has taught you that there is no true annihilation in particle physics, and in some sense the charged pions also “annihilate” in largely the same way, just with different intermediary states. There is no annihilation, but just the relentless force of entropy striving to spread energies more and more around the cosmos.
A: They are unstable because there are lighter particles they can decay into, consistent with the symmetries of the standard model.
You can well think of the quark and anti-quark annihilating.
