= Solution
We use these countability facts: the <integers> are <countable>; a finite <Cartesian product> of <countable sets> is <countable>; a <countable> union of <countable sets> is <countable>; and every subset of a <countable set> is <countable>. Finite products can be enumerated by listing index tuples in increasing maximum coordinate, and <countable> unions by diagonal enumeration of pairs of indices.
Fix $d\ge1$. For each degree bound $D$, there are finitely many monomials $X_1^{\alpha_1}\cdots X_d^{\alpha_d}$ with $\alpha_1+\cdots+\alpha_d\le D$. An <integer> <polynomial> of total degree at most $D$ is specified by an <integer> coefficient for each of them, so these <polynomials> form a finite <Cartesian product> of copies of $\mathbb Z$ and are <countable>. Every <polynomial> has some finite degree bound. Taking the union over $D=0,1,2,\ldots$ proves the <countability of integer polynomial rings>:
$$
\boxed{\mathbb Z[X_1,\ldots,X_d]\text{ is countably infinite}.}
$$
It is infinite because it contains all <integer> constant <polynomials>.
For $d=1$, each nonzero <polynomial> has only finitely many real roots: repeated application of the <factor theorem> bounds their number by its degree. The real <algebraic numbers> are the union of these finite root sets over the countably many nonzero <integer> <polynomials>. Thus they are <countable>. If the real <transcendental numbers> were <countable> too, their union with the <algebraic numbers> would make $\mathbb R$ <countable>, contrary to the allowed uncountability assumption. Hence \b[there are uncountably many real <transcendental numbers>].
To construct an <algebraically independent real sequence>, choose $x_1$ outside the <algebraic numbers>. Suppose $x_1,\ldots,x_d$ already satisfy the required nonvanishing condition. For each nonzero $P\in\mathbb Z[X_1,\ldots,X_d,T]$, write
$$
P(X_1,\ldots,X_d,T)=\sum_{j=0}^sP_j(X_1,\ldots,X_d)T^j.
$$
At least one coefficient <polynomial> $P_j$ is nonzero, and the inductive hypothesis makes its value at $(x_1,\ldots,x_d)$ nonzero. Consequently $P(x_1,\ldots,x_d,T)$ is a nonzero one-variable <polynomial> and has only finitely many real roots. There are countably many choices of $P$, so the union of all these forbidden roots is <countable>. Choose $x_{d+1}$ outside it, which is possible because $\mathbb R$ is uncountable. This extends the nonvanishing condition to $d+1$. Recursion therefore gives
$$
\boxed{f(x_1,\ldots,x_d)\ne0\quad\text{for every }d\ge1
\text{ and nonzero }f\in\mathbb Z[X_1,\ldots,X_d].}
$$
In particular the chosen numbers are distinct, since a relation $X_i-X_j$ is forbidden. Clearing rational denominators shows that every finite initial segment consists of <algebraically independent elements> over $\mathbb Q$.
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