Is this proof about the countability of $\Bbb Q \times \Bbb Q \times \cdots \times \Bbb Q$ sound? If $\Bbb{Q}$ is countable, prove that the set $\Bbb{Q}^n$ for $n = 2,3,...$ is countable.
Base case: $n = 2 \rightarrow \Bbb{Q}^2 = \Bbb{Q}\times\Bbb{Q}$ which, by Proposition 4.5 (see bottom of question), is countable.
Assume the statement is true for some $k \in \Bbb{Z}, k \ge 2,$ s.t., $\Bbb{Q}^{k}$ is countable.
Then $\Bbb{Q}^{k+1} = \Bbb{Q}^{k}\Bbb{Q}^{1}$ which is a relation $\Bbb{Q}^{k}\times\Bbb{Q}^{1}$. 
$\Bbb{Q}^{k}$ is countable by the inductive hypothesis and $\Bbb{Q}^{1}$ was introduced as a countable set. By Proposition 4.5, $\Bbb{Q}^{k+1}$ is a countable set.
By the PMI, $\Bbb{Q}^n$ is countable for $n = 2,3,...$ 
[Proposition 4.5: Let D be a countable set. Then D$\times$D (the set of all pairs $(x,y), x,y \in D$) is countable.]

As you wrote in a comment, the proposition you are trying to use doesn't apply directly to $\Bbb Q^k\times \Bbb Q$.
However, if we know that $D$ is countable and $E$ is some other countable set, then there is a bijection $f: D \to E$, and the map from $D \times D$ to $D \times E$ given by $(x,y) \mapsto (x,f(y))$ is a bijection. Since by the proposition $D \times D$ is countable, we have shown that $D \times E$ is countable too. 
This one you can apply to your proof, and the rest works as you wrote it.


Assume the statement is true for some $k \in \Bbb{Z},$ s.t., $\Bbb{Q}^{k}$

Are you missing something at the end of this sentence? I think it should read either "Assume the statement is true for some $k \in \Bbb{Z}$" or "Assume that  for some $k \in \Bbb{Z},$ s.t., $\Bbb{Q}^{k}$ is countable".
