If $C$ is a subset of $X$, then $C$ is bounded if and only if the absorbent hull of $C$ in $X$ is bounded. Now given an absorbent bounded subset $C$ of $X$, we may define a norm on its span so that $C$ is the unit ball. In short, we may find
a continuous map $W \to X$ where $W$ is a normed space and $C$ is the image of the unit ball in $W$. Conversely, the image under any continuous linear map from a normed space to $X$ of the unit ball in the domain is a bounded subset of $X$.
If $C$ is furthermore assumed closed, then $W$ will be a Banach space. Since the closure of a bounded set is again bounded, we may restrict attention to closed bounded sets. The existence of a countable basis of (closed) bounded sets in $X$
would then be equivalent to the existence of a collection of Banach spaces $W_n$ with continuous injections into $X$, such that any injection of a Banach space $W$ into $X$ factors through some $W_n$.
Now any Frechet space is ultrabornological, i.e. is the locally convex inductive limit of Banach spaces. Thus the existence of a countable basis of closed bounded subsets in $X$ would imply that $W$ is the locally convex inductive limit of a sequence of Banach spaces. A standard argument using the Baire category theorem would then imply that $W$ actually is a Banach space.
There are variants on the above argument, and one can probably make it more direct. One variant is as follows: if $X$ admits a countable basis of bounded sets, then the strong dual $X'_b$ admits a countable basis of seminorms, and so is again a Frechet space. But if the strong dual of a Frechet space is metrisable, then the original Frechet space is actually normable, and so is a Banach space. (If you unwind it, the proof of this latter statement is similar to the argument given above.)