How do we prove the existence of uncountably many transcendental numbers?

If a number $t$ is algebraic, it is the root of some polynomial with integer coefficients. There are only countably many such polynomials (each having a finite number of roots), so there are only countably many such $t$. Since there are uncountably many real (or complex) numbers, and only countably many of them are algebraic, uncountably many of them must be transcendental.

If you accept the premise that $|\mathbb R| = |P(\mathbb N)|$ then you know it is uncountable, due to Cantor's theorem.

Take all polynomials in rational numbers, each have only finitely many roots in $\mathbb R$.

Since the set of polynomials is equivalent to $\bigcup_{n\in\mathbb N} \mathbb Q^n$ (the union of polynomials from all degrees), which is countable you have only countably many possibly roots for all rational polynomials and thus only countably many algebraic numbers.

Now consider $\mathbb R\setminus A$ where $A$ is the set of all algebraic numbers, if it was not uncountable then $\mathbb R = \mathbb R\setminus A\cup A$ which was a union of two countable sets, which then would be countable.

There is a more number-theoretic way of looking at the matter, one that harks back to the work of Liouville.

He showed, among other things, that a number whose decimal expansion consists of $0$s and $1$s, such that the number of $0$s between consecutive $1$s grows faster than any polynomial, must be transcendental.

The standard example is the Liouville Constant, with the $n$-th gap being of length $n!$.

It is easy to produce continuum many transcendental numbers of the Liouville type.

The reason the result is worth mentioning is that it arises from examination of the approximability of algebraic numbers by rationals, a subject with a long and glorious history.

The required approximability result has a pretty quick proof that uses only high-school level mathematics.