Let me offer a simpler proof. First, let $c_{00}$ denote the subset of $\ell_2$ that consists of sequences with finite support, that is, sequences with finitely many non zero terms. Then your set, let's call it $r$, with rational entries, is a subset of this. I claim that $r$ is dense in $c_{00}$ and $c_{00}$ is in turn dense in $\ell_2$.
Proof Given $\varepsilon >0$ and $x\in c_{00}$, say $x=(x_1,x_2,\ldots,x_n,0,0,0,\ldots)$ pick rationals $r_1,\ldots,r_n$ such that $$|r_i-x_i|^2<2^{-i+1}\varepsilon$$
for $i=1,\ldots,n$. We can do this, for $\Bbb Q$ is dense in $\Bbb R$. Let, $q=(r_1,\ldots,r_n,0,\ldots)\in r$. Then we see $$\lVert x-q\rVert^2 =\sum_{i=1}^n|x_i-r_i|^2<\varepsilon\sum_{i=1}^n2^{-i+1}\leqslant \varepsilon\sum_{i=1}^\infty 2^{-i+1}=\varepsilon$$
This proves $r$ is dense in $c_{00}$. Now, we'll prove $c_{00}$ is dense in $\ell_2$. Pick thus $x\in\ell^2$. By definition, $$\sum_{n\geqslant 1}x_n^2<+\infty$$
where $x=(x_1,x_2,\ldots)$. In particular, given $\varepsilon >0$; there exists $N$ large enough so that $$\sum_{n> N}x_n^2<{\varepsilon}$$
But now all is easy: consider the element $x'=(x_1,\ldots,x_N,0,0,0,\ldots)\in c_{00}$. Then $$\lVert x-x'\rVert^2 =\sum_{n> N}x_n^2<{\varepsilon}$$
Thus $c_{00}$ is dense in $\ell_2$. Since density is transitive, we conclude $\bar r=\ell_2$, as desired. $\blacktriangleleft$
Observe this is easily adapted to prove $\ell^p$ is separable for each $p\geqslant 1$, for $r\subseteq c_{00}\subseteq \ell^p$ and $r$ is countable: it is essentially the same as $$\bigcup_{k\geqslant 1} \Bbb Q^k$$
As in the comments, we can take $$x^{(n)}=(\frac{1}{n},\frac{1}{n},\dots,\frac{1}{n},0,0,\dots)$$where the $n+1,n+2,...$-th entries of $x^{(n)}$ are all $0$. Then it needs to be checked that $x^{(n)}$ cannot converge in $d_{l_1}$ norm.
Edit.
Suppose that $x^{(n)}$ converges to some $x$ in $d_{l^1}$, then it can be checked that $x=(0,0,\dots)$ as follows:
Let $x(k)=$ the $k$-th coordinate of $x$. Then $$|x(k)-x^n(k)|\le d_{l^1}(x,x^{(n)})$$Taking $n\to\infty$ to see $x(k)=0$, hence $x=(0,0,\dots)$.
Best Answer
Let $(x_n)_{n \geq 1}$ be an arbitrary sequence in $\{0, 1\}$, so that $(x_n) \in X$. To prove the density of $Z$, we aim to show that there exists a sequence of elements of $Z$ that converge to $(x_n)$. This shows that $\bar{Z} = X$. Define the sequence, $(z_n^{(N)})_{n \geq 1} $ with,
$$z_n^{(N)} = x_n, \ \text{for} \ n \in \{1, \dots, N\} $$
$$z_n^{(N)} = 0, \ \text{for} \ n \geq N $$
So we have that,
$$(z_n^{(N)})_{n \geq 1} \in Z, \ \text{for every} \ N \geq 1 $$
We aim to prove that,
$$(z_n^{(N)})_{n \geq 1} \rightarrow (x_n)_{n \geq 1}, \ \text{as} \ N \to \infty $$
Since we investigate convergence in $(X, d)$ consider,
$$d((z_n^{(N)})_{n \geq 1}, (x_n)_{n \geq 1}) = \sum_{n=1}^\infty \frac{|x_n - z_n^{(N)}|}{2^n} = \sum_{n=N+1}^{\infty} \frac{|x_n|}{2^n}$$
Where the second equality comes from the fact that $z_n^{(N)} = x_n$ for $n \in \{1, \dots, N\}$, now we have that, using $|x_n| \leq 1$ for every $n$,
$$d((z_n^{(N)})_{n \geq 1}, (x_n)_{n \geq 1}) = \frac{1}{2^N} \sum_{n=N+1}^\infty \frac{|x_n|}{2^{n-N}} \leq \frac{1}{2^N} \sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2^N}$$
That is,
$$d((z_n^{(N)})_{n \geq 1}, (x_n)_{n \geq 1}) \leq \frac{1}{2^N}, \ \text{for every} \ N \geq 1$$
Taking $N \to \infty$ we obtain that,
$$d((z_n^{(N)})_{n \geq 1}, (x_n)_{n \geq 1}) \rightarrow 0, \ \text{as} \ N \to \infty$$
Therefore we have a sequence $(z_n^{(N)})_{n \geq 1} \in Z$ for every $N \geq 1$ that converges to $(x_n)_{n\geq 1} \in X$, therefore $\bar{Z} = X$.