Let X be a normed vector space and let $Y\subset X$ be a closed linear subspace. Suppose we have a sequence $(x_k)_{k\in \mathbb{N}}$ such that $$\inf_{y\in Y} ||x_k - x_{k+1} + y||_X < \frac{1}{2^k}, \quad \quad \text{for all} \ k\in \mathbb{N}.$$ Then by the Axiom of Countable Choice we can find a sequence $(\eta_k)_{k\in\mathbb{N}}$ in $Y$ such that $$||x_k - x_{k+1} + \eta_k||_X < \frac{1}{2^k}, \quad \quad \text{for all} \ k\in \mathbb{N}.$$
Why did we need to invoke the Axiom of Countable Choice here, surely it is obvious that we can just take the $y$'s such that the first expression above is satisfied, to be our $(\eta_k)$'s?
If not, what is the technical reason behind how the Axiom of Countable Choice works..how does it enable us to choose the sequence $(\eta_k)$?
Let's look at this from another perspective for a moment. Let's fix $x=x_k-x_{k+1}$, and ask what are the possible values of $y$ which satisfy $\|x-y\|<\frac1{2^k}$.
For a concrete example, consider $\Bbb R^2$ as $X$ and $\{(r,0)\mid r\in\Bbb R\}$ as $Y$. Now let $x=(0,0.01)$, and tell me how many possible values of $y$ we have to witness that $\inf_{y\in Y}\|x-y\|<\frac12$, for example. There are uncountably many possible values. You simply have no canonical choice of a vector.
So now you need to choose for every $k$ some point $\eta_k$ which witnesses that $x_k-x_{k+1}$ is sufficiently close to $Y$. And since there is no distinguished vector that just the job, there is no way of choosing these $\eta_k$'s uniformly. There is where the axiom of choice kicks in.
Your mistake is that when you write $\inf_{y\in Y}$, then $y$ is a quantified variable, and you can't treat it as a single valued notion. It ranges over all the possible values.