Non-existence of irrational numbers?

No. If you choose the digits independently and randomly, the resulting number is transcendental almost surely. (This follows from the fact that there are only countably many algebraic numbers, let alone rational numbers, and a countable subset of the reals has measure zero.)

You are confused about several things. Mathematicians have a precise and completely formal way of defining numbers that, in particular, does not depend on decimal expansions. Using these definitions, there are also precise and completely formal proofs that the resulting numbers are irrational or even transcendental.

This is the problem Pythagoras faced. We once believed all numbers could be expressed as a ratio of two integers, hence the term rational number. The diagonal of a unit square is $\sqrt 2$ which is irrational. This is easy to see. Take two unit squares and cut them along their diagonals. You now have four right triangles whose legs are each equal to $1$. Arrange them into a square by putting the four right angles together. Since this one square is made from two unit squares, the whole thing has area equal to $2$. Therefore, the sides of the square which correspond to the diagonals of our original unit squares have length of $\sqrt{2}$. Therefore, this number is real. You may like neither limits nor non-repeating decimals, but irrational numbers exist.

Let me answer your question with another question. What rational number contains ever sequence in its decimal expansion? It turns out none. Either the decimal terminates or it has some point where it becomes periodic (a terminating decimal could be seen as 0 repeating).