Why is an infinitely large capacitor a DC open circuit?
Did you ever pick up a telescope, hold the wrong end to your eye, and find you can't make sense of what you see? 

The only thing wrong with using this telescope to see how an infinite capacitance behaves

\$I=C\frac{dU}{dt}\$

is that you aren't holding the useful end to your eye.

Try it this way round

\$\frac{dU}{dt}=\frac{I}{C}\$

What this equation means is that you can shove any finite amount of current, however large, into an infinite capacitor, and its voltage will not change. This is a convenient approximation for the behaviour of 'very large' capacitors, whose voltage 'does not change much' during operation. LTSpice has an infinite capacitor, and I would guess most other spices do as well. These are useful for approximating coupling or decoupling capacitors, when you are not interested in their deviation from ideal.

Now if you rephrase your question as 'I have a component whose terminal voltage cannot change, and I change its terminal voltage ...', well, you get what you get.

You might want to edit your title so that says 'DC short circuit'.