Vtyjkyuk

In Naive Set Theory by Halmos, there is an assertion given as follows:

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 $n$.

In the preceding paragraph, he gives a definition of the number of elements in a finite set as:

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 $mathcalP(x)$ to $omega$.
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'm trying to prove the assertion.

I have proven the auxiliary step via mathematical induction on $n$ as follows.

To prove: if $m, n in omega$, then $(m + n) - n sim n$

1. (induction base) $n = 0$

beginalign
& quad (m + 0) - m & \
&= m - m & text(by definition of addition) \
&= emptyset & \
&= n sim n
endalign

2. (induction hypothesis) $(m + n) - m sim n$

(induction conclusion) $(m + n^+) - m sim n^+$

beginalign
& quad (m + n^+) - m & \
&= (m + n)^+ - m & text(by definition of addition) \
&= (m + n) cup m + n - m & text(by induction hypothesis)\
&= [(m + n) - m] cup m + n
endalign

By induction hypothesis, there exists a bijection $f$ from $(m + n) - m$ to $n$.
We can construct a bijection $g$ from $[(m + n) - m] cup m + n$ to $n^+$ as follows:

$$g(x) =
begincases
f(x) & textif x in (m+n) - m \
n & textif x = m + n
endcases
$$

and thus, $(m + n^+) - m sim n^+$.

Now, to prove the assertion I suppose that it's crucial to prove the second sentence about the disjoint sets, from which it would be obvious that the union of two finite sets is finite.

I don't understand however how to start with that proof.
Maybe I should consider a resulting set $(E cup F)$ and construct some bijection from it to some other set (possibly a natural number), or use the auxiliary fact to somehow connect relationship between the "arbitrary" sets in a problem $E, F$ and the equivalent natural numbers?