Since by definition, $\displaystyle \mathbb{Z}[X_1,...,X_d] = \mathbb{Z}[X_1,...,X_{d-1}][X_d]$ it remains to prove that if $\displaystyle A$ is countable then $\displaystyle A[X]$ is countable.Show $\displaystyle \mathbb{Z}[X_1,X_2,...,X_d]$ is countable.
Let $\displaystyle A_n[X] = \{ f(x) \in A[x] : \deg f(x) = n \}$ for $\displaystyle n\geq 0$.
Thus, $\displaystyle A[x] = \bigcup_{n=0}^{\infty} A_n[x]$.
Also, $\displaystyle |A_n[x]| = |A^n|$ is clear by defining a bijection.
Since each ($\displaystyle n\geq 0$) $\displaystyle A^n$ is countable.
We have that $\displaystyle A[x]$ is a countable union of countable sets.
Thus, it is countable.
.Show there are uncountably many transcendental numbers
Given an algebraic number $\displaystyle \alpha$ it needs to satisfy a non-zero polynomial over $\displaystyle \mathbb{Q}$. There are countably many non-zero polynomials over $\displaystyle \mathbb{Q}$ and each one has finitely many zeros. Since every algebraic number is among those polynomials and conversely a zero of one of those polynomials is algebraic it follows there are countably many algebraic numbers. Thus, there have to be uncountably many transcendental numbers. Because $\displaystyle |R| > |N|$.