@DanielFischer So is the following right?

$Y$ is dense in $\ell^2(\mathbb{N})$.

This means that for every $x$ there is a sequence $(y_n)$ in $Y$ such that $\lVert y_n - x\rVert \to 0$.

We have $\inf \{ \lVert x-y\rVert : y \in Y\} \leqslant \lVert x - y_n\rVert$ for every $n$, since $y_n \in Y$. But $\lVert x - y_n\rVert \to 0$ implies $\inf \{ \lVert x-y\rVert : y \in Y\} = 0$.

But it cannot be that there is a $y \in Y$ such that $||x-y||=0 \Leftrightarrow x=y$ since $x \notin Y$.

Answer:
$${n-1 \choose k-1} - k {n-x-1 \choose k-1}$$