Suppose that $A$ is a simple C*-algebra (i.e. there is no closed ideal $I\subset A$ such that $0\neq I\neq A$) and let $(H,\phi)$ be a representation. Can we conclude that $(H,\phi)$ is faithful, i.e. that $\phi$ is injective?
Since $\ker(\phi)$ is a closed ideal in $A$, we must have $\ker(\phi)=\{0_{A}\}$ or $\ker(\phi)=A$, by simplicity of $A$. But how do we exclude the case $\ker(\phi)=A$?
If this is not true for general representations, what can we say about the injectivity of irreducible representations in this setting?
You exclude the case $\ker\phi=A$ by not allowing the dimension of $H$ to be zero. Allowing a Hilbert space of dimension zero (so equal to $\{0\}$) gives you nothing, and it ruins several things. Its only orthonormal basis is $\varnothing$, for instance.
Let us mention some problems with $B(H)$ for the Hilbert space $H=\{0\}$:
$B(H)=\{0\}$, the zero operator, which is also the identity operator
every operator is injective
every operator is surjective
as every operator is invertible, you have $\sigma(0)=\varnothing$. So the spectrum is empty for every operator (not that there is more than one).
In summary, allowing Hilbert spaces of dimension zero will give you absolutely nothing useful, and at the same will ruin the statements of many theorems.