Halmos proves shortly before the cited paragraph that finite subsets are not equivalent to themselves. He then says the following:
The number of elements in a finite set E is, by definition, the unique natural number equivalent to E; we shall denote it by #(E). It is clear that if the correspondence between E and #(E) is restricted to the finite subsets of some set X, the result is a function from a subset of the power set $\mathcal{P}(x)$ to $\omega$.
This is all clear (a specification on dom[$E \rightarrow \#(E)$] of $s : s \subset X$). He then continues:
This function is pleasantly related to the familiar set-theoretic relations and operations. Thus, for example, if $E$ and $F$ are finite sets such that $E \subset F$, then $\#(E) \leq \#(F)$. (The reason is that since $E \cong \#(E)$ and $F \cong \#(F)$, it follows that $\#(E)$ is equivalent to a subset of $\#(F)$.)
I do not follow his reasoning. The fact itself is clear, but I do not see the implication which he is suggesting. Is he indicating that $E$, as a subset of $F$, will be mapped by this function to some natural number which is a subset of $\#(F)$?
Edit: The passage continues:
Another example is the assertion that if $E$ and $F$ are finite sets, then $E \bigcup F$ is finite, and, moreover, if $E$ and $F$ are disjoint, then $\#(E \bigcup F) = \#(E) + \#(F).$ The crucial step in the proof is the fact that if $m$ and $n$ are natural numbers, then the complement of $m$ in the sum $m+n$ is equivalent to $n$; the proof of this auxiliary fact is achieved by induction on $n$. Similar techniques prove that if $E$ and $F$ are finite sets, then so also are $E \times F$ and $E^F$, and, moreover, $\#(E \times F) = \#(E) * \#(F) $ and $\#(E^F) = \#(E)^{\#(F)}$.