Proof of $\ell_\infty$ is not separable

Question

Let $\ell_\infty=\{\bar{x}=(x_1,x_2,x_3,\dots) \,:\,x_n\in \mathbb{R}, \sup_{n\in\mathbb{N}} |x_n|<\infty, n\in \mathbb{N} \}$ and $$d_\infty(\bar{x},\bar{y})=\sup_{n\in\mathbb{N}} |x_n-y_n|,$$ for all $\bar{x},\bar{y} \in \ell_\infty$. Then, $(\ell_\infty, d_\infty)$ is not separable.
I want to prove $(\ell_\infty, d_\infty)$ is not separable by showing that there exists a subset of $\ell_\infty$ which is countable but not dense in $\ell_\infty$.
In the class, my lecture proved it in a similar way as in the question of this post: Show that $l_\infty$ is not separable.
I want to try with another subset of $\ell_\infty$.
This is my attempt.
Given a set of sequences $Y=\{\bar{y}=(y_1,y_2,\dots,y_n,0,0,0,\dots)\, :\, y_k\in \mathbb{Q}, 1\leq k\leq n, n\in\mathbb{N}\}$.
Hence, $Y$ is a countable set.
Then, take any $\bar{y}\in Y$, so $\bar{y}=(y_1,y_2,\dots,y_n,0,0,0,\dots)$ where $y_1,y_2,\dots,y_n\in \mathbb{Q}$. Since the sequence $\bar{y}$ is finite, then $$\sup_{n\in \mathbb{N}} |y_n|<\infty.$$ So, $\bar{y}\in \ell_\infty$.
Thus, $Y$ is a countable subset of $\ell_\infty$.
Next, I will show that $Y$ is not dense in $\ell_\infty$.
Choose $\varepsilon_0=\frac{1}{4}$ and $\bar{x}=(1,1,1,1,\dots)\in \ell_\infty$. Take any $\bar{y}\in Y$, then \begin{align*} d_\infty(\bar{x},\bar{y})&=\sup_{n\in \mathbb{N}} |x_n-y_n|\\ &=\sup \{|x_1-y_1|,|x_2-y_2|,\dots,|x_n-y_n|,|x_{n+1}-y_{n+1}|,|x_{n+2}-y_{n+2}|,\dots \}\\ &=\sup \{|1-y_1|,|1-y_2|,\dots,|1-y_n|,|1-0|,|1-0|,\dots \}\\ &=\sup\{|1-y_1|,|1-y_2|,\dots,|1-y_n|,1,1,\dots \}\\ &\geq \sup\{1-|y_1|, 1-|y_2|,\dots,1-|y_n|,1,1,\dots \}\\ &=1>\frac{1}{4}=\varepsilon_0. \end{align*} Because $d_\infty(\bar{x},\bar{y})\geq\varepsilon_0$, so $Y$ is not dense in $\ell_\infty$.
Hence, $(\ell_\infty, d_\infty)$ is not separable.
Is there any mistakes of my proof?
Thanks for any help.