Is it possible to construct a strictly monotonic sequence of all rational numbers? I know that the set of all rational numbers is countable, and can be enumerated by a sequence, say $\{a_n\}$.  But can we construct a monotonic $\{a_n\}_{n=1}^{\infty}$, e.g. with $a_k<a_{k+1}$?  It doesn't seem plausible to me, because then $a_1$ would be the smallest rational number, which clearly can't be any finite number.  Am I mistaken?

As stated, the answer is no, because the question uses the symbol $<$ which has the implied meaning: The usual ordering of $\mathbb{Q}$ where $\frac{a}{b}<\frac{c}{d}$ iff $ad < bc$ in $\mathbb{Z}$. 
But.
As mentioned in another answer, $\mathbb{Q}$ can be well-ordered, i.e. one can define a different order $\prec$ with the property that every nonempty subset of $\mathbb{Q}$ contains a least element with respect to $\prec$. For this ordering, a monotone sequence containing all of the rationals is easy to construct: let $x_1$ be the smallest rational, let $x_2$ be the smallest element of $\mathbb{Q} \setminus \{x_1\}$, etc.
