I'm trying to walk through the standard proof of uncountability of the reals. I assume there is a surjection $f: \mathbb{N} \to \mathbb{R}$, and for every $f(n)$, I write out their decimal expansions: \begin{align*} f(0) & = a_{00} . a_{01} a_{02} a_{03} \\ f(1) & = a_{10} . a_{11} a_{12} a_{13} \\ f(2) & = a_{20} . a_{21} a_{22} a_{23} \\ & \vdots \end{align*} and then construct a real number not in the image. In the proof I have in mind, there are two steps I take which I believe require choice:
(a) When I write out the decimal expansions of $f(n)$, I want to say, "if it doesn't have a unique decimal expansion, choose the one with an infinite string of zeroes rather than an infinite string of $9$'s."
(b) When I define $b = b_0 . b_1 b_2 \ldots$, I want to say for every $j$, choose $b_j \neq a_{jj}$ and $b_j \neq 9$. I am fairly sure this requires the axiom of choice.
In each step, I'm making a finite choice, but I'm doing it infinitely many times. Am I correct that both of these steps require the axiom of choice?
Neither of these requires the axiom of choice. You can explicitly write down the choice function in both cases: in the first case you already have, and in the second case you can choose $b_j$ to be the smallest possible admissible digit. This is a proof of the uncountability of $\mathbb{R}$ in ZF.
You didn't ask about this, but it is also completely unnecessary to assume that $f$ is a surjection; you make no use of that hypothesis in the proof. The proof simply constructs, for every function $f : \mathbb{N} \to \mathbb{R}$, a real number not in the image of $f$, which proves that no such function can be surjective. There is no need to invoke proof by contradiction; this is a direct proof that $\mathbb{R}$ is uncountable.