If the supply is current limited, the answer is simple. Same if you have a fuse. However ... but I assume your question is directed at supplies that do not, in which case things become a little more interesting....
5 amps is not a huge current in itself. Yes, it is at the high range of the sort of thing you might encounter on your average bench, but it is by no means scary.
A 10V supply delivering 5A means the load will be equivalent to 2 ohms. If the supply is rated at these specifications, then Ideally) it should be able to deliver that current indefinitely. Expect some parts to heat up, but they
should all operate within their capabilities. (In practice, this is not always the case - but let's not dwell on that right now.)
Now, increase the load by putting, say, 1 ohm across the output. You are now trying to draw 10A. This is a (significant) overload.
In this situation, the ratings of the output transistors may be exceeded and/or their heatsinking may be inadequate. Failure is on the cards. In the case of a linear supply (big, heavy transformer) the voltages within the unit will almost certainly sag as the transformer is incapable of supplying the demand. In the case of a switchmode unit, you could get shutdown or hiccuping. In either case, power and current ratings of components within a unit that is trying to work will be at risk of being exceeded. Things will generally get hot and you are very likely to have something "give".
Now let's move onto the "short".
Let's say you stick a screwdriver across the output terminals, presenting an effective load of, say, 0.1 ohms. This would have the supply try to deliver 100A ... which just isn't going to happen - and something is going to fail. It's a similar process as the overload mentioned above, but is usually much quicker.
The obvious answer - the dead short is a load that consumes infinite amps.
This is the direction my points were leading to, but infinite amps requires zero resistance and that would require a superconductive current path. However, for a theoretical discussion of an ideal system, this is correct.