Character on Banach Algebra is Surjective/Proof 1

Proof
From Image of Submodule under Linear Transformation is Submodule, $\phi \sqbrk A$ is a vector subspace of $\C$.

From Dimension of Proper Subspace is Less Than its Superspace, we have:
 * $\dim \phi \sqbrk A \le \dim \C = 1$

and so we either have $\phi \sqbrk A = \set 0$ or $\phi \sqbrk A = \C$.

Since $\phi \ne 0$ by the definition of a character, we have $\phi \sqbrk A = \C$.