According to the phase diagram at http://www.lsbu.ac.uk/water/phase.html, if you compress it slowly enough (so that the temperature remains constant at 300 K), the water will compress slightly and slightly more as you increase the pressure to about 1 GPa (10,000 atm), at which point it will have a density of about 1.18 g/cm$^3$ (18% greater than usual).
At this point, if you continue to apply more pressure, the water will
suddenly solidify into begin forming a crystal known as ice VI. This does not have the same structure as ordinary ice (ice I), which makes sense because ordinary ice is less dense than water, so it will never be formed by compression. Ice VI, on the other hand, is more dense than water (about 1.31 g/cm$^3$), and that's what you get if you apply a pressure greater than 1 GPa.
If you increase the pressure even more, the ice VI changes phase a few more times (forming ice VII, then ice X, it looks like), before getting to the limit of pressures available in the lab.
If you somehow managed to apply much, much, more pressure than this (for example by throwing the water into a neutron star), then kakemonsteret's answer would begin to apply. The water would first become electron-degenerate matter (like a white dwarf), then the nuclei would start to fuse and it would become like neutron star material. But the pressures required for this are many orders of magnitude higher than anything available outside of a compact star, and at such pressures the steel in your experiment would suffer a similar fate.