This turns out not to be the case.

Let's play this game exactly once.

You choose two unequal real numbers. I don't know what they are, and I don't know the distribution from which you choose them. You write them down and put them in separate envelopes.

I'm allowed to choose one envelope and open it to see the number inside, and my job is then to say which envelope holds the larger number.

I claim I have a strategy now which lets me win strictly more than 50% of the time. My strategy is this.

I choose a real number R at random from a distribution that has dense support. In other words, for any two reals, L and U with L<U, P(L<R<U) > 0. This is easy to do ... one method is to list the rationals, positive and negative, then roll a die, discarding numbers until you get a 6.

Now I flip a coin and thereby choose an envelope at random. I proceed by assuming my chosen number is between your two numbers. There is a non-zero chance this is true ... call it e. So e>0.

If I'm wrong then my choice is 50% ... probability is 1-e.

If I'm right then my choice is 100%. ... probability is e.

Combined, my chance of being right is 0.5(1-e) + e = 0.5+e/2, which is strictly greater than 50%.

You can make it as small as you like, and if we play the game repeatedly then you can make it approach 50%. But as it stands, with a one-off game, I can win with a probability that depends on your chosen numbers, but which is strictly bigger than 50%.

Yes, this two real numbers question suffers from hidden conditions as the original version of Monty Hall problem does, but even more explicitly.

We can't have an uniform distribution over all real numbers, so it's quite pointless to discuss if looking into the envelope gives any new information, cause we don't even know the distribution yet.