Why don't molecules of a gas settle?
You are used to all collisions being somewhat lossy – that is, when you think of most collisions, a little bit of the kinetic energy is lost at each collision so the particles will slow down. If they are subject to gravity, they will eventually settle.
By contrast, the collisions between gas molecules are perfectly elastic – for a non-reactive gas (mixture), there is no mechanism by which the sum of kinetic energies after the collision is less than before. *
Even if an individual gas molecule briefly found itself at rest against the bottom of the container, the thermal motion of the molecules of the container would almost immediately give it a "kick" and put it back into circulation.
There is a theorem called the equipartition theorem that tells us that for each degree of freedom, the gas molecules will on average contain $\frac12 kT$ of energy. This is an average – individual molecules may at times have more or less. But the average must be maintained – and this means the gas molecules keep moving.
One way you can get the molecules of the gas to settle at the bottom of the container would be to make the walls of the container very cold – taking thermal energy away, the molecules will eventually move so slowly that the effect of gravity (and intermolecular forces) will dominate. That won't happen by itself – you need to remove the energy somehow.
To estimate the temperature you would need: for a container that is 10 cm tall, the gravitational potential energy difference of a nitrogen molecule is $mgh = 1.67\cdot 10^{-27}\ \mathrm{kg}\cdot 28 \cdot 9.8\ \mathrm{m\ s^{-2}}\cdot 0.1\ \mathrm m= 4.6\cdot 10^{-26}~\mathrm{J}$. Putting that equal to $\frac12 kT$ gives us a temperature of
$$T = \frac{2 m g h}{k} = 3.3\ \mathrm{mK}$$
That's millikelvin. So yes – when things get very, very cold, gravity becomes a significant factor and air molecules may settle near the bottom of your container.
* Strictly speaking, this is a simplification. With sufficient energy, some collisions can lead to electronic excitation and even ionization of the molecules. The de-excitation of these states can result in radiative "loss" of energy, but if the system is truly closed (perfectly isolated) the radiation will stay inside until it's re-absorbed. Still, this means that, at least for a little while, kinetic energy may appear to be "lost". Similarly, there are some vibrational modes for molecules that get excited at sufficiently high energy/temperature; in these modes, energy moves from "kinetic" to "potential" and back again – so that it is not "kinetic" for a little bit of the time.
An important consideration in all this is "what is the temperature of anything that the gas can exchange energy with". That is not just the walls of the container (although their temperature is very important), but also the temperature of anything the gas "sees" – since every substance at non-zero temperature will be a black body emitter (some more efficiently than others), if the gas can exchange radiation with a cooler region, this will provide a mechanism for the gas to cool. And if the gas gets cold enough, gravity wins.
In fact, particles in a box of gas are slightly denser at the bottom than they are at the top. In general, the probability of finding a particle with a total energy of $E$ is proportional to the Boltzmann factor: $$ P(E) \propto e^{-E/kT}. $$ In particular, the potential energy of a gas molecule is $mgh$, where $h$ is the height above some fixed point (the "floor" of the box, say.) If we consider the relatively probabilities of a particle to be found at the floor of a box ($h = 0$) versus being found at a height $h$ above the bottom of a box, we will have $$ \frac{P(h)}{P(0)} = \frac{e^{-mgh/kT}}{e^{0}} = e^{-mgh/kT}. $$ Thus, the densities of the gas molecules is lower at a height $h$ than it is near the floor, since they are less likely to be found at these heights.
The problem is that this factor is tiny for typical temperatures and masses of gas molecules. For the air in my office, we have $m \approx 32$ amu (the mass of an oxygen molecule, $h \approx 3$ m (the height from floor to ceiling), and $T \approx 293$ K. Plugging these all in, we get that the density of the air at the ceiling is 99.961% of the density at the floor.
Based on the various comments, I think it's worth noting that even molecules in a solid don't "settle" — they too are constantly in rapid motion.
The difference between a solid and a gas is that, in the solid, each molecule is confined to a small volume through interactions with nearby molecules.
When a solid appears to "settle", what's really happening is that all of the obvious macroscopic motion simply gets converted into making the individual molecules vibrate faster within their confines. (with some energy lost due to transferring the momentum to the surrounding air particles)