Let $|\alpha \rangle$ be coherent state in Fock space. According to the paper "Coherent-state representation for the photon density operator" by Cahill (Phys. Rev. 138, B1566 (1965), §VII), every convergent series $\{\alpha_j\}$ on the complex plane generates a complete set $\{|\alpha_j \rangle \}$ of the Fock space. That means, every vector $| \psi \rangle$ can be written as super position $| \psi \rangle = \sum_j g_j |\alpha_j\rangle$.
That also means, that any coherent state can be written as $$ | \alpha \rangle = \sum_j c_j |\alpha_j \rangle . $$ Let us suppose that $|\alpha \rangle \notin \{|\alpha_j \rangle \}$. That means at least two coefficients $c_j$ are non-zero. Now, let us apply the annhilation operator $a$ to the coherent state. By definition we get $a |\alpha \rangle = \alpha |\alpha \rangle$. Using the expansion, we get $$ a | \alpha \rangle = \sum_j c_j \alpha_j |\alpha_j \rangle \stackrel{!}{=} \alpha \sum_j c_j |\alpha_j \rangle . $$ Doesn't that imply, that $\alpha_j = \alpha$, which contradicts our early assumption, that $|\alpha \rangle \notin \{|\alpha_j \rangle \}$? Or it means that the set $\{|\alpha_j \rangle \}$ is overcomplete, which means the super position is not unique. But is there a way to remove elements, to make it a non-overcomplete bases such that the super position is unique? Wouldn't we then get a contradiction as indicated above?