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.