what's wrong about following proof? It's obviously wrong as we never used finite dimension of $\mathcal{H}$...
Lemma: Let $A$ be C*-algebra. Let $\pi: A \to \mathcal{B(H)}$ be irreducible *-representation and $\text{dim}\mathcal{H}=n\in \mathbb{N}$. Then $\pi$ is surjective.
(wrong) proof: The commutant $\pi(A)'$ of $\pi(A)$ is equal to $\mathbb{C}\cdot I$ (property of irreducible representations. $I$ is identity operator). Hence $\pi(A)''=\mathcal{B(H)}$. From Von neumann double commutant theorem we get $\overline{\pi(A)}=\pi(A)''=\mathcal{B(H)}$. But $\pi(A)$ is closed (as homomorphic image of *-algebra isomorphism) in $\mathcal{B(H)}$, hence $\pi(A)=\mathcal{B(H)}$.