If $X$ is separable then so is $B_X.$ 
Let $X$ be a separable normed linear space. Then show that $B_X = \{x \in X\ |\ \|x\| = 1\}$ is also separable.

Since $X$ is separable it has a countable dense subset. Let $D = \{x_n\ |\ n \in \Bbb N \}$ be a countable dense subset of $X.$ WLOG we may assume that $0 \not\in D$ for otherwise we could take $D \setminus \{0\}$ which continues to be dense in $X.$ Then I guess that $D' = \left \{\frac {x_n} {\|x_n\|}\ \bigg |\ n \in \Bbb N \right \}$ is a dense subset of $B_X.$ Let $y \in X$ with $\|y\| = 1.$ Let us choose $\varepsilon \gt 0$ arbitrarily. Then I need to show that $B(y,\varepsilon)$ intersects $D'.$ If $y = x_n$ then we are through. If not then there exists a sequence $\{x_{k_n}\}_{n \geq 1}$ in $D$ converging to $y.$ Since norm is a continuous function it follows that $\|x_{k_n}\| \to \|y\| = 1$ as $n \to \infty.$ But this shows that $\frac {x_{k_n}} {\|x_{k_n}\|} \to y$ as $n \to \infty,$ proving that $D'$ is dense in $B_X.$
Does my proof make sense? Can anybody plaese check my argument above?
Thanks in advance.

A shorter proof could be : If $X$ is metric and separable then has a countable basis. Any subspace of a countable basis space is countable basis (take the trace on the subspace), and in the end recall that if a space has a countable basis is separable.
