Here we show that every normed space that has a Schauder basis is separable. Note that we are (implicitly) assuming that the normed spaces in question are spaces over the field $K$ where $K$ is either $\mathbb{R}$ or $\mathbb{C}$ . So let $(X, \left\|\cdot\right\|)$ be a normed space with Schauder basis, say $S = \left\{e_1, e_2,\dots\right\}$ . Notice that our notation implies that $S$ is infinite. In finite dimensional case, the same proof with a slight modification will yield the result.

Now, set $Q$ to be the set of all finite sums $q_1e_1+\cdots + q_ne_n$ such that each $q_j = a_j + b_ji$ where $a_j, b_j \in \mathbb{Q}$ . Clearly $Q$ is countable. It remains to show that $Q$ is dense in $X$ .

Let $\epsilon > 0$ . Let $x \in X$ . By definition of Schauder basis, there is a sequence of scalars $(\alpha_n)$ and there exists $N$ such that for all $n \geq N$ we have, $$\left\|\sum_{j=1}^{n}\alpha_je_j - x\right\| < \epsilon/2$$ But then in particular, $$\left\|\sum_{j=1}^{N}\alpha_je_j - x\right\| < \epsilon/2$$ Furthermore, by density of $\mathbb{Q}$ in $\mathbb{R}$ , we know that there exist constants $a_1,\dots, a_N, b_1,\dots, b_N$ in $\mathbb{Q}$ such that, $$\left\|\sum_{j=1}^N(a_j + b_ji)e_j - \sum_{j=1}^N\alpha_je_j\right\|<\epsilon/2$$ By triangle inequality we obtain: $$\left\|\sum_{j=1}^N(a_j + b_ji)e_j - x\right\| \leq \left\|\sum_{j=1}^N(a_j + b_ji)e_j - \sum_{j=1}^N\alpha_je_j\right\| + \left\|\sum_{j=1}^{N}\alpha_je_j - x\right\| < \epsilon$$ Noting that $$\sum_{j=1}^N(a_j + b_ji)e_j$$ is an element of $Q$ (by construction of $Q$ ) and that $x$ and $\epsilon$ were arbitrary, we conclude that every neighborhood of $x$ contains an element of $Q$ , for all $x$ in $X$ . This proves that $Q$ is dense in $X$ and completes the proof.