= Solution
Map each bounded sequence $a=(a_n)$ to the continuous function that takes value $a_n$ at the integer $n\geq1$ and is affine on every interval $[n,n+1]$. Convexity of absolute value gives $|F_a(t)|\leq\sup_n|a_n|$, and evaluation at integers gives equality of suprema. Thus $a\mapsto F_a$ is an <isometric embedding> of the <l-infinity sequence space> into $C^b([1,\infty))$.
The former is not <separable> by the allowed fact. If the latter were <separable>, the subspace result just proved would make its isometric copy of $\ell_\infty$ <separable>, a contradiction. Hence
$$
\boxed{C^b([1,\infty))\text{ is not separable}.}
$$
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