Definitions and theorems I'm working with
Recursion theorem: If $A$ is a set, $f:A\to A$ is a function and $a\in A$ then there exists a unique function $g$ s.t. $$g\left(0\right)=a$$$$g\left(n+1\right)=f\left(g\left(n\right)\right)$$
A set is finite if it is bijectable with some $\left[n\right]$ where $n\in\mathbb{N}$. A set is infinite if it is not finite.
Proposition: If $A$ is an infinite set then $\left|A\right|\geq\aleph_{0}=\left|\mathbb{N}\right|$
Proof: Let $f$ be a choice function on $\mathcal{P}\left(A\right)$, define a recursive $g:\mathbb{N}\to A$ with $g\left(0\right)=a=f\left(A\right)$ and $g\left(n+1\right)=f\left(A\backslash\left\{ g\left(k\right)\mid k<n\right\} \right)\notin\left\{ g\left(k\right)\mid k<n\right\}$ which is clearly injective (By induction) and well defined since at any point we cannot have $A\backslash\left\{ g\left(k\right)\mid k<n\right\} =\emptyset$ as this would mean $\left|A\right|=\left|\left[k\right]\right| $ and contradict $A$ being infinite.
Putting aside minor details (such as the omitted induction), is this valid? I ask because the proof I have in my lecture notes is much more complicated, and I think needlessly so.
Yes, this is the standard straightforward proof.
With a more involved proof, however, it is possible to get this result from only the axiom of Countable Choice rather than needing unrestricted choice. Without knowing more about your lecture notes, my first guess would be that that's what's going on there.
(To use only countable choice, choose for each $n\in\mathbb N$ a subset of $A$ with $n$ elements and a concrete bijection of that subset with $\{0,1,\ldots,n-1\}$. Then you can prove that the union of the subsets is countably infinite without making more choices).