I wanted to find a quick proof of the following well-known fact. Since I couldn't easily find a reference, I provide a proof below.
Let $H$ be a separable Hilbert space, and $J\subset B(H)$ be a (not necessarily closed) two-sided ideal. If $T\in J$ is non-compact, then $J=B(H)$.
I'm also interested to know if there is a shorter/simpler argument.
By using the polar decomposition, we can write $T=V|T|$. So $|T|=V^*T\in J$, and then $J$ contains a positive non-compact operator. On a side note, this argument also shows that $J$ contains all adjoints of its operators, since now $T^*=|T|V^*\in J$.
So from now on we assume $T\geq0$, non-compact, $T\in J$. This means that there is $\lambda>0$ with $\lambda\in \sigma(T)$ such that the spectral projection of $T$ corresponding to $(\lambda-\varepsilon,\lambda+\varepsilon)$ is infinite for all $\varepsilon>0$. By choosing $\varepsilon=\lambda/2$, we get an infinite projection $P$ such that $TP=PT$ and $$TP\geq(\lambda/2)P.$$ This means that, in the C$^*$-algebra $PB(H)P$, the positive operator $TP$ is invertible: there exists $S\in B(H)$ such that $(PSP)TP=P$. But as $T\in J$, this implies that $P\in J$.
Now fix an orthonormal basis $\{e_j\}$ of $H$ and an orthonormal basis $\{f_j\}$ of $PH$. Since both have the same cardinality (here is where we use that $H$ is separable), there exists an isometry $W\in B(H)$, given by $We_j=f_j$, such that $WW^*=P$. Then $$ I=W^*W=W^*WW^*W=W^*PW\in J. $$ Thus $J=B(H)$.