Imagine we put some volume of water in an infinitely strong container, and then start to slowly exert a force on the top of the water , with increasing pressure. What happens to the water over time? It would certainly begin to compress, but to what extent? would it turn into ice at some point?
Water Compression – What Happens to Water as It is Increasingly Compressed?
fluid-staticsphase diagramphase-transitionpressurethermodynamics
Related Solutions
There's actually not one simple answer to your question, which is why you are a bit confused. To specify your problem fully, you must specify exactly how and whether the gas swaps heat with its surroundings and how or even whether it is compressed. You should always refer to the full gas law $P\,V=n\,R\,T$ when reasoning. Common situations that are considered are:
Charles's Law: The pressure on the volume gas is constant. No work is done by the gas on its surroundings, nor does the gas do any work on its surroundings or piston or whatever during any change. The gas's temperature is that of its surroundings. If the ambient temperature rises / falls, heat is transferred into / out from the gas and its volume accordingly increases / shrinks so that the gas's pressure can stay constant: $V = n\,R\,T/P$; with $P$ constant, you can retrieve Charles's Law;
Isothermal: the gas is compressed / expanded by doing work on / allowing its container to do work on its surroundings. You think of it inside a cylinder with a piston. As it does so, heat leaves / comes into the gas to keep the temperature constant. As the gas is compressed, the work done on it shows up as increased internal energy, which must be transferred to the surroundings to keep the temperature constant. At constant temperature, the gas law becomes $P\propto V^{-1}$;
Adiabatic: No heat is transferred between the gas and its surroundings as it is compressed / does work. AGain, you think of the gas in a cylinder with a plunger. This is prototypical situation Feynman talks about. As you push on the piston and change the volume $V\mapsto V-{\rm d}V$, you do work $-P\,{\rm d}V$. This energy stays with the gas, so it must show up as increased internal energy, so the temperature must rise. Get a bicycle tyre pump, hold your finger over the outlet and squeeze it hard and fast with your other hand: you'll find you can warm the air up inside it quite considerably (put you lips gently on the cylinder wall to sense the rising temperature). This situation is described by $P\,{\rm d}V = -n\,\tilde{R}\,{\rm d} T$. The internal energy is proportional to the temperature and the number of gas molecules, and it is negative if the volume increases (in which case the gas does work on its surroundings). But the constant $\tilde{R}$ is not the same as $R$: it depends on the internal degrees of freedom. For instance, diatomic molecules can store vibrational as well as kinetic energy as their bond length oscillates (you can think of them as being held together by elastic, energy storing springs). So, when we use the gas law to eliminate $P = n\,R\,T/V$ from the equation $P\,{\rm d}V = -n\,\tilde{R}\,{\rm d} T$ we get the differential equation:
$$\frac{{\rm d} V}{V} = - \frac{\tilde{R}}{R}\frac{{\rm d} T}{T}$$
which integrates to yield $(\gamma-1)\,\log V = -\log T + \text{const}$ or $T\,V^{\gamma-1} = \text{const}$, where $\gamma=\frac{R}{\tilde{R}}+1$ is called the adiabatic index and is the ratio of the gas's specific heat at constant pressure to the specific heat at constant volume.
The example you have chosen to start your question (water) is an atypical case. The phase diagram has a logarithmic scale for the pressure, somewhat hiding the negative slope of the transition line between the liquid and solid phases. That negative slope says that by increasing pressure, a solid in a state close to the melting curve will melt, and a liquid also close to the phase transition line will remain liquid.
However, by further increasing pressure, liquid water eventually solidifies. The reason is that the negative slope of the melting curve is connected to the lower density of normal ice Ih relative to the liquid. This is possible if the crystalline structure is open with low coordination. At high pressure, more dense solid phases with higher coordination appear, thus making a negative slope of the melting curve over the whole phase diagram unlikely.
The relation between the slope of the melting curve in the $P-T$ plane and differences in entropy $(\Delta s$ and molar volume $(\Delta v$ at the transition is ruled by Clapeyron's equation : $$ \frac{{\mathrm d}P}{{\mathrm d}T} = \frac{s_{liq}-s_{sol}}{v_{liq}-v_{sol}}. $$ The numerator on the right-hand side can be rewritten as $\frac{{\cal l}}{T}$, where ${\cal l}$ is the latent heat of the liquid-solid transition (positive). Therefore, the sign of $v_{liq}-v_{sol}$ controls the sign of the melting curve.
As far as I know, by increasing the pressure at a constant temperature, eventually, all materials have a fluid-crystal transition. This is consistent with the idea that by increasing pressure, the solid becomes more and more compact, making it unlikely that at very high pressure $v_{liq}-v_{sol}<0$. A final word of caution is about the possibility that, in some cases, kinetic effects may influence the practical observability of the transition to a crystalline solid.
Best Answer
This can be qualified by a phase diagram. This shows the most stable state of matter at a given combination of pressure and temperature (or other variables, but $p-T$ is most common).
The phase diagram of water, as can be found on Wikipedia, looks like this:
If we assume that temperature is constant during the process (maybe you apply the pressure slowly, so that the system can always equilibriate with the surroundings), increasing the pressure corresponds to going vertically upwards on the phase diagram.
If you start at room temperature and 1 atm, you have liquid water -- but if you compress it enough (to about 1 GPa), it becomes a solid!
You can also see that the solid water you will produce will not be the normal ice from your freezer (which is ice $I_h$), but an exotic phase of ice called ice VI. And applying even more pressure will convert it to other phases of ice, VII, X and XI. The differences between those solid phases are their crystal structure. Unlike liquid and gas, which can only have one phase (because the molecules can move around), a solid can comprise multiple phases. The atoms stay in one place (more or less), so their arrangement matters.
But because you proposed an indestructible vessel, you seem to be interested in really high pressures. I have not found what can happen to water, but stuff seems to be doing some really wierd things -- just check out metallic hydrogen. And the common salt NaCl seems to form some interesting structures, some of which can also behave as metals. So maybe, at some point, one could get metallic water...
And that is not the end. By compressing water to extremely high densities (or any other fermionic matter), you encounter the Pauli exclusion principle -- the electrons simply cannot get too close to each other. At this point, you will get something akin to a white dwarf. Compressing even more, the electrons and protons will combine to form neutrons and you will get a "neutron star"-like state. Even more pressure, and you end up with a black hole. Such states of matter are called degenerate. And that will truly be the end.