Ross' Elementary Analysis - 11.7 Let $(r_{n})$ be an enumeration of the set $\mathbb{Q}$ of all rational numbers. Show that there exists a subsequence $(r_{n_{k}})$ such that $\lim_{k \to \infty} = +\infty$. 
Theorem If the sequence $(s_{n})$ is unbounded above, it has a subsequence with limit $+\infty$.
Please critique or vote on the proof I supply as a response. Thanks!
 A: Proof. Since $\mathbb{Q}$ is unbounded above, by the theorem above, we know that $(r_{n})$ contains a subsequence that diverges to $+\infty$. 
Lemma: $\mathbb{Q}$ does not have an upper bound. 
Proof of Lemma. Suppose for contradiction that $\mathbb{Q}$ does have an upper bound. That is, there is some $q$ such that for all $x \in \mathbb{Q}$, we have $q \geq x$. 
First consider some positive $z \in \mathbb{Q}$, and fix some corresponding $r \in \mathbb{Q}$ such that we know $z \geq \frac{q}{r}$. Notice here that $q$ must be positive since 100000000 is a member of $\mathbb{Q}$ and $q$ must be greater than it. 
We have two cases to consider: either $r$ is positive or $r$ is negative. Suppose $r$ is positive; we know that $zr \geq q$ which is a contradiction since $zr \in  \mathbb{Q}$ due to multiplicative closure.
Suppose there is no positive $r$ such that $z \geq \frac{q}{r}$. That is for all $r>0$, $z < \frac{q}{r}$. Consider then $r=-1$ in which case $z > q$ contradicting our original claim as required.
