Why is the image of a C*-Algebra complete?

To expand on my comment, let us take as given the following two (nontrivial) facts:

Lemma 1: If $U$ is a $C^*$-algebra and $I\subset U$ is a closed 2-sided $*$-ideal, then $U/I$ is a $C^*$-algebra.

Lemma 2: If $U$ and $V$ are $C^*$-algebras and $\pi:U\to V$ is an injective $*$-morphism, then $\|\pi(A)\|=\|A\|$ for all $A\in U$.

Now suppose $\pi:U\to V$ is any $*$-morphism of $C^*$-algebras; we wish to show $\pi(U)$ is closed. Let $I$ be the kernel of $\pi$; then $\pi$ factors through an injective $*$-morphism $\pi':U/I\to V$. By Lemma 1, $U/I$ is also a $C^*$-algebra, so we can apply Lemma 2 to $\pi'$ and conclude that $\pi'$ is an isometry. In particular, this implies the image of $\pi'$ is closed. But the image of $\pi'$ is the same as the image of $\pi$, hence the image of $\pi$ is closed.

Let me now briefly sketch the proofs of the lemmas. In Lemma 1, it suffices to show that the usual quotient norm satisfies the $C^*$ identity. This can be shown by some calculations using an approximate identity for $I$ (the idea being that the best way to approximate $A\in U$ by elements of $I$ is to take $AB_\alpha$, where $B_\alpha\in I$ is an approximate identity for $I$). In Lemma 2, you can use the $C^*$ identity to reduce to the case that $A$ is self-adjoint, in which case the norms of $A$ and $\pi(A)$ can be computed as the spectral radius. It thus suffices to show that $\pi$ preserves spectra of self-adjoint elements. This can be shown using the continuous functional calculus (if $\sigma(A)$ were strictly larger than $\sigma(\pi(A))$, you could find two continuous functions on $\sigma(A)$ that agree on $\sigma(\pi(A))$, and this contradicts injectivity of $\pi$).