Solution: Let $X$ be a normed linear space and $\mathcal{B} = \{ e_n \} $ be Schauder basis of $X$. We need to prove $X$ is separable, that is, it has a countable dense set. For $n\in \mathbb{N} $, we define \[ A_n \coloneqq \left\{ \sum_{i=1}^{n} q_i e_i : q_i \in \mathbb{Q} \right\} . \] Then we have a natural bijection from $\mathbb{Q} ^n$ to $A_n$ given by \[ f: \mathbb{Q} ^n \to A_n,\quad (q_1, q_2,\dots, q_n) \mapsto \sum_{i=1}^{n} q_i e_i. \] This shows that each $A_n$ is countable. Thus the set \[ A \coloneqq \bigcup _{n\in \mathbb{N} } A_n \] is also countable (countable union of countable sets is countable).

We claim that $A$ is dense in $X$. Let $x\in X$. Since $\mathcal{B} $ is a Schauder basis of $X$, we can find a sequence of real numbers $(c_n)$ such that \[ \sum_{i=1}^{n} c_i e_i \rightarrow x. \] So, given any $\varepsilon > 0$ we can find $N\in \mathbb{N} $ such that \[ \left\lVert \sum_{i=1}^{N} c_i e_i - x \right\rVert \lt \frac{\varepsilon }{2}. \] Since, $\mathbb{Q} $ is dense in $\mathbb{R} $, for $1 \leq i \leq N$, there exists $q_i \in \mathbb{Q} $ such that \[ \left\vert q_i - c_i \right\vert \lt \frac{\varepsilon }{2 N \left\lVert e_i \right\rVert }. \] Thus we have \begin{align*} \left\lVert \sum_{i=1}^{N} q_i e_i - x \right\rVert & = \left\lVert \sum_{i=1}^{N} (q_i - c_i)e_i + c_i e_i - x \right\rVert \\ & \leq \left\lVert \sum_{i=1}^{N} (q_i - c_i) e_i\right\rVert + \left\lVert \sum_{i=1}^{N} c_i e_i - x \right\rVert \\ & \leq \sum_{i=1}^{N} \left\vert q_i - c_i \right\vert \left\lVert e_i \right\rVert + \left\lVert \sum_{i=1}^{N} c_i e_i - x \right\rVert \\ & \lt \sum_{i=1}^{N} \frac{\varepsilon }{2N} + \frac{\varepsilon }{2} = \varepsilon . \end{align*}