For any Hilbert space $H$, one can show that the ideal of compact operators, is closed with respect to the strong operator topology by using that the unit ball is weakly compact or directly in the case when $H$ is separable using a diagonal argument.
The cantor diagonal argument fails when trying to see if the ideal of compact operators, is closed in $B(H)$ with respect to the weak topology. For this reason I have tried to find a counter example showing that $B(H)_{\text{compact}}$ is not closed in $B(H)$.
In the case when $H=:L_2(\mathbb{T})$, we have the sequence $T_n$ of elements in $B(H)_{\text{compact}})$ given by $T_n(f)=\mathscr{F}^{-1} \widehat{f(*+n)}$, i.e. the shift in fourier space. $T_n$,a sequence of compact operators, converges in the weak operator topology to $0$ by testing against all trigonometric polynomials.
The only problem with this example is that it is not a nonexample!: The zero map is a compact operator.
I am convinced however, that one can use an idea like this to show that the space of compact operators in $B(H)$ is not closed.
Question: How do I do it?
The space $\mathcal{K}(H)$ of compact operators is not closed with respect to the strong operator topology (SOT) of $H$ is infinite dimensional. Take $H = \ell^2$, and let $\{e_n\}$ be the standard orthonormal basis of $H$. Then, for any $x\in H$, we may write $$ x = \sum_{n=1}^{\infty} \langle x,e_n\rangle e_n $$ So if $P_n(x)$ denotes the $n^{th}$ partial sum of this series, then each such $P_n$ is bounded and has finite rank, and is thus compact. Also, $P_n(x) \to x$ for each $x\in H$, so $P_n \to I$ strongly, and $I\notin \mathcal{K}(H)$.