Let $c$ denote the normed space of convergent sequences of real numbers with norm $$||(x_1, x_2, x_3, \cdots)|| = \sup \{ |x_n| : n \in \mathbb{N} \}.$$
Note that $c$ is a Banach space. To show that $c$ is separable, we will show that there exists a countable subset of $c$ that is dense in $c$. Consider the set $$S := \{ (x_1, x_2, x_3, \cdots) \mid x_1 = x_2 = \cdots \text{ and } x_i \in \mathbb{Q} \}$$
Thus $S$ is the set of all constant sequences with rational limit. Note the countability of $\mathbb{Q}$ implies the countablity of $S$. We claim that $S$ is dense in $c$. To see that $S$ is dense in $c$, let $$(a_1 , a_2 , \cdots ) \in c$$
Let $\sup \{ a_n \mid n \in \mathbb{N} \} = a \in \mathbb{R}$. Let $\varepsilon > 0$. By the density of $\mathbb{Q}$ in $\mathbb{R}$, there exists an $x \in \mathbb{Q}$ such that $x \in (a – \frac{\varepsilon}{100}, \: a + \frac{\varepsilon}{100}$). Now define $(p_1 , p_2 , \cdots ) = (x,x, \cdots) \in S$. Now, we have that $$||(a_1 , a_2 , \cdots ) – (p_1 , p_2 , \cdots )|| < \varepsilon.$$
Thus $(a_1 , a_2 , \cdots ) \in \bar{S}$. So $S$ is a countable dense subset of $c$ as desired. $\square$
Above is my attempt at proving $c$ is separable. Please let me know if my proof is correct or if there are places for improvement.
Best Answer
Your proof is not correct, note that $$ \| a - p \|_\infty = \sup_n |a_n - p_n| = \sup_n |a_n - x| $$ will in general be quite large. You have chosen $x$ such that $x $ is close to $\sup_n a_n$, but if $(a_n)$ is not almost constant, this does not help. For example, look at $a = (\frac 1n)$, then $\sup_n \frac 1n = 1$, so you can choose $x = 1$. But then $\|a - p\|_\infty = \sup_n 1- \frac 1n = 1$, which is not $\epsilon$-small.
Here is a hint to guide you to improve your attempt: The above shows that constant sequences are not enough, but almost constant sequences will do. That is, look at $$ S = \{x \in c : \text{there is $N$, such that $x_n$ is constant for $n \ge N$}, x_n \in \mathbf Q, \text{ all $n$} \} $$ Then $S$ is countable (can you prove that?). To show that $S$ is dense, argue along the following line: Let $x\in c$, $\epsilon > 0$ be given. As $x$ is convergent, say to $l$, choose $q \in \mathbf Q$ such that $|l - q| < \frac\epsilon 2$. There is $N$ such that $$ |x_n - q| \le |x_n - l| + |l - q| < \epsilon, \quad n \ge N $$ For $n < N$ choose $q_n \in \mathbf Q$ such that $|x_n - q_n| < \epsilon$. Then $s := (q_1, \ldots, q_{N-1}, q, q, \ldots) \in S$ is $\epsilon$-close to $x$.