Showing a C* Algebra contains a compact operator In my functional analysis class we are currently dealing with C* Algebras, and I just met this problem:

Let $ \mathbb{H} $ be a separable Hilbert space, and suppose we have $ A \subset B(\mathbb{H}) $ a C* Algebra of bounded operators of bounded operators on H. Now we suppose there exists $ a \in A $ and that there exists a compact operator $ K \in K(\mathbb{H}) $ such that $ ||a-K|| < ||a|| $. We are to show A contains a compact operator on H that is not the zero operator, that is $ A \cap K(H) \neq \{0\} $.

I am quite new to C* Algebras and I still have not much intuition but I cannot really see how to do this. I cannot seem to find a way of doing this. I certainly appreciate all help.

Here is an answer in terms of a standard result of C*-algebra theory.

Theorem: Every injective $*$-homomorphism from one C*-algebra to another isometric.

In addition to this result, we will need to know that the quotient of a C*-algebra by a closed ideal makes sense, and that the result is a C*-algebra. In particular, we have the Calkin algebra $Q(H)=B(H)/K(H)$, the C*-algebra of bounded operators modulo the compact operators. 
Consider now the restriction to your algebra $A$ of the quotient map $\pi:B(H) \to Q(H)$. Supposing that your algebra $A$ contains no compact operator, this restriction is injective, hence isometric by the theorem. That is, for every $a \in A$, the norm of $a$ equals the norm of $\pi(a) \in Q(H)$. But, recalling the definition of the quotient norm, this says exactly that
$$\|a\| = \inf_{k \in K(H)} \|a-k\|$$
i.e. the distance from $a$ to the compacts equals the norm of $a$. Therefore, there is not any $k \in K(H)$ with $\|a\| > \|a-k\|$
It would probably be an instructive exercise to unpack the proof of the theorem and see how it applies to this particular case.
