Algebra of compact operators is CCR algebra In Arveson's book "Invitation to $C^*$-algebra's", it is claimed that every algebra of compact operators is a CCR algebra.
Concretely, let $\mathcal{A}$ be a $C^*$-subalgebra of some $B_0(\mathcal{H})= \mathcal{K}(\mathcal{H})$ where $\mathcal{H}$ is a Hilbert space.
Let $\pi: A \to B(\mathcal{K})$ be an irreducible representation.
Why is $\pi(A) \subseteq B_0(\mathcal{K})$?
 A: What Arveson claims is that 

(Theorem) 1.4.4 implies that every C$^*$-algebra of compact operators is CCR. 

So you have that $\pi=\sum_{j\in J}\pi_j$ where each $\pi_j$ is irreducible, the family is orthogonal, and each $\pi_j$ is equivalent to a subrepresentation of the identity representation of $A$. 
Since $\pi$ is irreducible, the orthogonality of the family implies that it has cardinality 1: if $\pi_1$ and $\pi_2$ are pairwise orthogonal representations of $A$, then the support projections $P_1,P_2$ of $\pi_1,\pi_2$ are in $\pi(A)'$, so $\pi$ is not irreducible. 
Thus $\pi$ is equivalent to a subrepresentation of the identity representation of $A$. This means that there exists a projection $P\in B(H)$ and a unitary $U:PH\to K$ such that $\pi(a)=UPaPU^*=UaU^*$ for all $a\in A$. 
Now consider a projection $q\in A$. If $\pi(q)=UqU^*$ is not compact, it is of infinite-rank. So there is a strictly decreasing sequence of projections $q\geq q_1\geq q_2\geq\cdots$. Now $\{U^* q_jU\}$ is a strictly decreasing sequence of projections in $A$. As these are compact, they are finite-rank, so we get a contradiction. It follows that $\pi(q)$ is finite-rank. So all projections in $\pi(A)$ are finite-rank, and so every elements in $\pi(A)$ is compact. 
