The space of bounded continuous functions are not separable

Consider the subset $K$ of $C_b(\Bbb R)$ consisting of functions that are either $0$ or $1$ at the integers. There is an uncountable subset $S$ of $K$ such that: $$\tag{1}\Vert x-y\Vert\ge 1,\ \ \text{whenever}\ \ x,y\in S\ \text{with}\ x\ne y.$$

Now, given a countable subset $B$ of $C_b(\Bbb R)$ it follows from $(1)$ that there is an $s\in S$ that is distance at least $1/2$ from every element of $B$. Then $B$ is not dense in $C_b(\Bbb R)$.

Consider the maps $f_t(x):=\exp(itx)$. We have, if $t_1\neq t_2$, that \begin{align*} |f_{t_1}(x)-f_{t_2}(x)|&=|\exp(it_1x)-\exp(it_2x)|\\ &=\frac 12\left|\exp\left(i\frac{t_1+t_2}2x\right)\right|\cdot \left|\exp\left(i\frac{t_1-t_2}2x\right)-\exp\left(-i\frac{t_1-t_2}2x\right)\right|\\ &=\sin\left(\frac{t_1-t_2}2x\right), \end{align*} which gives, when $t_1\neq t_2$, $$\lVert f_{t_1}-f_{t_2}\rVert_\infty=1.$$

Here is also one way, incase you know that $\ell^{\infty}$ is not separable, the set of bounded sequences with sup-norm that is.

Define $\Lambda:C_{b}(\mathbb{R})\to\ell^{\infty}$ by $f\mapsto (f(n))_{n=1}^{\infty}$. By choosing any $(x_{n})_{n=1}^{\infty}\in \ell^{\infty}$ we can define a continuous bounded function $f\in C_{b}(\mathbb{R})$ by setting $f(n)=x_{n}$ for all $n\in\mathbb{N}$ and extending it to $\mathbb{R}$ by connecting the imagepoints of each natural number linearly. We may take $f(x)= x_{1}$ for all $x<1$, for example. Since $f\mapsto (x_{n})_{n=1}^{\infty}$, then $\Lambda$ is a surjection. It is also continuous, because \begin{equation*} \|\Lambda(f)-\Lambda(g)\|_{\infty}=\sup_{n\in\mathbb{N}}|f(n)-g(n)|\leq \sup_{x\in\mathbb{R}}|f(x)-g(x)|=\|f-g\|_{\infty} \end{equation*} for all $f,g\in C_{b}(\mathbb{R})$.

Now if $C_{b}(\mathbb{R})$ was separable, it would have a countable and dense subset $\mathscr{D}$. Since the image of a dense subset under a continuous surjection is dense, we have that $\Lambda(\mathscr{D})$ is a countable, dense subset of $\ell^{\infty}$. This would imply that $\ell^{\infty}$ is separable, which is a contradiction.