Show that there are infinitely more problems than we will ever be able to compute

Question

I was looking at this reading of MIT on computational complexity and on minute 15:00 Erik Demaine embarks on a demonstration to show what is stated in the title of this question. However I cannot follow his reasoning, in practice what he says is this:
we can state a decision problem as string of $0$ and $1$ which in practice is the truth table of the function.
He goes on to say that a decision problem is an infinite string of bits while a program is a finite string of bits and up to here no problem. What I don't understand is the continuation of the proof from this point on: Decision problems are in $\mathbb{R}$ cause you can put a decimal point before the string representing the problem, thus obtaining the decimal part of a real

 for example if you have 0111011010101010101... it could become x.0111011010101010101... 

A program is "just" an integer in $\mathbb{N}$ cause it is a finite string of bits. The point that I fail to understand is how it is possible that a decision problem is comparable to a real number instead of an integer ... I mean, if we use the argument of "put a dot in front of the number" could not the same reasoning also be applied to the number of possible algorithms that can ever be produced?