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One of my homework problems asks to prove that if $f:X\to Y$ is a function and $A$ is a countable subset of $X$, then $f(A)$ is countable.

I believe I have a valid proof by partitioning the set A into equivalency classes based in image under $f$. Then using that partition as domain for a choice function $C_f$ back to $A$. So now the function $f$ restricted to the domain (range $C_f$) $f|_{range(C_f)}$ maps to $f(A)$ 1-1 and onto. Therefore

$$|f(A)|=|range(C_f)|\leq |A|=\aleph_0$$ $\therefore f(A)$ is a countable set. $\quad \blacksquare$

My problem is that I used the axiom of countable choice $AC_\omega$ in my proof but that isn't a part of the course I am taking. Does anyone have a simpler proof or one that does not involve Choice?

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You have indeed invoked choice, but not in an essential way.

Remember that by assumption $A$ is countable - so you have a map $g$ from $A$ to $\mathbb{N}$. This gives us a well-ordering of $A$: $$\alpha\prec\beta\iff g(\alpha)

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    Thank you! I don't know why I didn't notice that before. I have a small error in my proof. I am given $A$ is countable but not necessarily infinite. So $|A|\leq \aleph_0$ – 2017-02-27