The CNO process does not end in carbon. It's part of the hydron burning processes. Let's look at the different processes inside main sequence stars:
Hydrogen burning describes the fusion process from the lightest possible element (hydrogen) into the 2nd lightest element (helium). Or if you look at it at the level of protons and neutrons it converts 4 separate protons into a one nucleus which consists of 2 protons and 2 neutrons. The difference in binding energy is released in this process.
Now, if you go into the details even further, there are different ways this process can happen as it is not a single reaction but involves several intermediate steps which may be different and occur with a different likelyhood but which all result in essence in the same.
proton-proton-chain (pp-chain):

Let's look at the processes which occur there:
- happens twice: $p + p \rightarrow {}^2H + \nu_e + e^+ $
This is the limiting reaction as it requires a weak interaction (it converts a proton into a neutron). It takes on average for a single proton 9 billion years inside the Sun. We are lucky there are so many protons inside the Sun.
- happens twice: ${}^2H + p \rightarrow {}^3He + \gamma$
This reaction is comparatively quick as it is based on the strong nuclear force (no conversion of protons into neutrons, just nucleus transformation) and estimated to happen within one second after the generation of the ${}^2H$.
- ${}^3He + {}^3He \rightarrow {}^4He + p + p$
In (1) you gain additional radiation energy of 1.442MeVif you consider the annihilation of the positron with a free electron which will occur immediately after. This the so-called ppI branch. If the collision of the ${}^3He$ was with one of the existing (and here produced) ${}^4He$ nuclei we would get the (in the Sun less likely) ppII and ppIII chains.
So overall we have the equation of $4{}^1_1H + 2e^- \rightarrow {}^4_2He + 2\nu_e + 7\gamma + 26.7MeV$
Now the CNO cycle (or also called Bethe-Weizsäcker-cycle in honour of the two people who suggested its existence independently and simultaneously) which effectively yields the same and also is a series of reactions. I will only illustrate the main cycle here:

- ${}^{12}_6C + {}^1_1H \rightarrow {}^{13}_7N + \gamma$ (strong force)
- ${}^{13}_7N \rightarrow {}^13_6C + e^+ +\nu_e$ (weak force, but decay, ~10 minutes)
- ${}^{13}_6C + {}^1_1H \rightarrow {}^{14}_7N + \gamma$ (strong force)
- ${}^{14}_7N + {}^1_1H \rightarrow {}^{15}_8O + \gamma$ (strong force)
- ${}^{15}_8O \rightarrow {}^{15}_7N + e^+ + \nu_e$ (*weak force, but decay, ~2 minutes)
- ${}^{15}_7N + {}^1_1H \rightarrow {}^{12}_6C + {}^4_2He$ (strong force)
Now, the weak force must not be overcome here anywhere in fusion but only plays a role on larg(er) instable nuclei which decay into nuclei, thus is not the time-limiting factor. Yet, it requires a proton to "collide" with a much heavier nucleus which already combines 6 (C) or 7 (N) protons along with some neutrons. Much more energy is needed to overcome this coulomb barrier. Thus this CNO cylce is strongly temperature (and pressure dependent):

And this is the reason that in the sun (and other dwarf stars) the pp process is the dominant one. But this process does not yield any other material than ${}^4_2He$, especially it does not yield $C$, $N$ or $O$ which are only used as catalyst; they are neither effectively produced nor consumed; their presence is a mere requirement for this process to happen and thus this process cannot have happened in the first stars (which only consisted of hydrogen and helium) in their main sequence regardless of mass and pressure.
Note also that the CNO cycle does only rarely create a (stable) ${}^{16}_8O$, but there ist most often only ever the non-stable ${}^{15}_8O$ which decays within 2 minutes.
In the last step, theoretically (and with a small probability) a ${}^{16}_8O$ could be produced, but this is energetically less favourable.
However, even then the ${}^{12}C$ is not permanently
destroyed, because except in about one out of $5\cdot 10^7$ cases, ${}^{16}O$ will again return to ${}^{12}C$ (cf. $8)
(quoted from Bethe's original publication)
Carbon is produced only via the triple alpha process. It happens when the amount of helium in a star's core becomes so little that the likelyhood of a hydrogen fusion sinks and density and temperature rose enough to start the triple alpha process which fuses helium into carbon. This requires MUCH higher pressure and densities than hydrogen fusion as you need to bring together three particles of twice the charge than previously within $10^{-16}s$ as ${}^8_4Be$ decays quickly and its creation is endothermic. The star leaves the main sequence phase at this point. The other alpha processes to create higher-order elements can start when enough carbon has been produced by the triple alpha process.