$\mathcal{A}+K$ is norm-closed where $\mathcal{A}$ is a $C^*$-algebra and $K$ is the compact operators. Let $\mathcal{A}\subset B(H)$ be a unital $C^*$-algebra and let $K$ be the closed ideal of compact operators. I need to show that $\mathcal{A}+K$ is also a $C^*$-subalgebra of $B(H)$. I am stuck at proving that the space $\mathcal{A}+K$ is norm-closed. 
I begin with a a sequence $\{B_n\}_{n=1}^{\infty}$ in $\mathcal{A}+K$ with $B_n\rightarrow B$, for some operator $B\in B(H)$. Now, for each $n\in\mathbb{N}$, $B_n=A_n+K_n$, where $A_n\in \mathcal{A}$ and $K_n\in K$. And from here I'm lost. Is there some kind of Bolzano-Weierstrass property for compact operators? Hints are welcome. 
 A: Note that $\mathcal{A}\cap K$ is a closed ideal in $\mathcal{A}$, and we have a well-defined $\ast$-homomorphism
$$
\varphi : \mathcal{A}/(\mathcal{A}\cap K) \to B(H)/K \text{ given by } a + (\mathcal{A}\cap K) \mapsto a+K
$$
The range of $\varphi$ is precisely $(\mathcal{A} + K)/K$, and the range of a $\ast$-homomorphism is complete. 
Since $K$ is complete, it follows that $\mathcal{A} + K$ is also complete; and hence closed.
Edit: Added is a proof of the following fact: 

If $A$ is a normed linear space and $B\subset A$ a closed subspace such that $B$ and $A/B$ are both complete, then $A$ is complete

Suppose $(x_n) \subset A$ is Cauchy, since $\|(x_n + B) - (x_m+B)\| \leq \|x_n - x_m\|$, it follows that $(x_n+B)$ is Cauchy in $A/B$. By hypothesis, $\exists y\in A$ such that
$$
x_n + B\to y+B
$$
By definition of the quotient norm, for each $n\in \mathbb{N}, \exists z_n \in B$ such that
$$
\|x_n - y - z_n\| < \|(x_n+B) - (y+B)\| + 2^{-n}
$$
So if $y_n := x_n - y - z_n$, then
$$
y_n \to 0 \text{ in } A
$$
Hence, $z_n = x_n - y - y_n$ is Cauchy in $B$ (being the difference of two Cauchy sequences). Since $B$ is complete, $\exists z\in B$ such that
$$
z_n \to z
$$
and hence
$$
x_n = (x_n - y-y_n) + (y+y_n) \to z+y
$$
