My professor approved my approach to proving that $\phi$ is multiplicative using a counting argument and a bijection. Since we do not cover rings I could not take the easier and more general approach using $\mathbb{Z_n}$ (mostly because I have yet to study those topics). He mentioned my proof is fine for this class but he said if I wanted to be extra pedantic I should mention why the function I introduced is well defined.
Here is my proof:
We claim that $\phi(mn)=\phi(m)\cdot\phi(n)$, when $\gcd(m,n)=1$. To show this is true we will apply a counting argument. The goal is to construct two sets of integers, one that has $\phi(mn)$ elements and the other $\phi(m)\phi(n)$ elements. If we can establish a bijection between the sets then we will have shown that $\phi(mn)=\phi(m)\phi(n)$.
We begin by constructing the set $Z_1$ defined as $$Z_1=\{a\mid1\leq a\leq mn\textrm{ and } \gcd(a,mn)=1\}$$ That is $Z_1$ contains all the positive values less than or equal to $mn$ and coprime to $mn$. By definition the size of this set is $\phi(mn)$. Also let us define the set $Z_2$ as $$ Z_2=\{(b,c)\mid1\leq b\leq m\textrm{ and }\gcd(b,m)=1 \textrm{ and }1\leq c\leq n\textrm{ and }\gcd(c,n)=1\}$$ We choose the first value of the ordered pair namely $b$ which can be done in $\phi(m)$ ways. Then for the second value of the ordered pair namely $c$ which can be done in $\phi(n)$ ways. Then by definition of cardinality the total number of ordered pairs that can be constructed in $Z_2$ is the product of the two values, that is $\phi(m)\phi(n)$ ways. To show the sets $Z_1,Z_2$ are the same cardinality we will define a bijection $f: Z_1\rightarrow Z_2$. for each $a\in Z_1$ define $f(a)=(b,c)$ where $(b,c)\in Z_2$, with $b\equiv a\pmod{m}$ and $c\equiv a\pmod{n}$. To see that $f$ is injective, let $f(a)=f(b)$, then by definition we have $a\equiv b\pmod{m}$ and $a\equiv b\pmod{n}$. Then we know by the Chinese Remainder Theorem, $a\equiv b\pmod{mn}$. Since $a,b\in Z_1$ it follows that $a=b$. To show $f$ is surjective consider the congruences $x\equiv b\pmod{m}$ and $x\equiv c\pmod{n}$. Again by the chinese remainder theorem these two congruences have a simultaneous solution $a$ that is unique modulo $mn$. That is we have found an $a$ such that $f(a)=(b,c)$. This completes the proof and the statement that $\phi$ is multiplicative is true.
I am unsure by what he means by well-defined as I thought the Chinese remainder theorem gives us that 1-to-1 correspondence.