Define of completely continuous operator : $L$ is continuous operator and map bounded set to relatively compact set , then $L$ is completely continuous operator.
Let $\Omega$ is a bounded set of Banach space and $L$ is completely continuous. Is $L(\Omega)$ finite dimensional set ?
No. Here's a prototypical example. I'll leave details of proof.
Suppose that $H$ is a Hilbert space with an orthonormal basis $(\phi_n)$ and let $(\lambda_n)$ be a sequence of numbers converging to $0$. Define
$$T f = \sum_{n=1}^{\infty} \lambda_n (\phi_n,f) \phi_n.$$
Then $T$ is compact (the image of any bounded set is relatively compact).
Note that the range of $T$ is finite dimensional if and only if all but finitely many $\lambda_n$'s are zero. In fact, any self-adjoint compact operator on $H$ is of this form.
I'd like to add that your intuition was in the right direction, though, as $T$ is the limit in the operator norm of the sequence of the finite-rank operators $T_N$ obtained by taking only the first $N$ summands, and more generally, the linear space of compact operators is the closure of the finite rank operators.