Let us imagine some algebraic equation of order $n$. Clearly, this alegraic equation can be determined by its coefficients. Thus, we denote some algebraic equation by the Cartesian product of all of its coefficients, $(a_{n-1}, \ ..., \ a_1, \ a_0)$. For algebraic equations of order $n$, the set of all possible coefficient products is the set: $$A_n \ = \ \displaystyle\prod_{i \ = \ 1}^{n} \ \mathbb{Q}$$ We know that $\mathbb{Q}$ is countable (from exercise 1), and we know that a finite product of countable sets is countable, therefore $A_n$ for any $n \ \in \ \mathbb{Z}$ is countable. Let $A \ = \ \{A_1, \ A_2, \ ... \}$. One can check that $A$ is countable quite easily, as its elements are indexed by the positive integers. We know that a countable union of countable sets is countable, meaning that: $$S \ = \ \displaystyle\bigcup_{n \ \in \ \mathbb{Z}} A_n$$ is also countable. Let us pick some element $s \ \in \ S$. This element $s$ will have a finite set of algebraic numbers associated with it, which we call $X_s$ (as we assumed at the beginning of the theorem). Since $S$ is countable, the union of all $X_s$, which we call $X$, is a countable union of countable sets (since each $X_s$ is finite). By definition, $X$ is the set of algebraic numbers. Thus, the set of algebraic numbers is countable.

Assume that the set of transcendental numbers, $\textbf{T}$, is not uncountable (and is thus countable). By definition, $\textbf{T} \ = \ \mathbb{R} \ - \ \textbf{A}$, where $\textbf{A}$ is the set of algebraic numbers. Thus, $\textbf{T} \ \cup \ \textbf{A} \ = \ \mathbb{R}$. A countable union of countable sets is countable, so by our initial assumption and Part A, this would imply that $\mathbb{R}$ is countable, which is a contradiction. Thus, $\textbf{T}$ must be uncountable.