Let $H$ be a infinite-dimensional separable Hilbert space and $\mathcal{I}$ be a proper closed two-sided ideal of $B(H)$. Can $\mathcal{I}$ contain a projection for a infinite dimensioal proper closed subspace? If $H$ is not separable (or $\mathcal{I}$ not closed) while all other conditions remain the same, will the conclusion change? What if $\mathcal{I}$ is only one-sided (other conditions remain the same)?
(Added) This question is inspired by Corollary 5.11 in Banach Algebra Technique in Operator Theory written by Ronald G. Douglas. This corollary claims that in an infinite dimensional separable Hilbert Space the ideal of compact operators $\mathcal{K}$ is the only proper closed two-sided ideal in $B(H)$. The proof first assumes that $T \in \mathcal{I}$ and $T$ is NOT compact. Hence by the converse of Lemma 5.8 in the range of $T$ there is a closed infinite dimensional subspace say $M$. Define $ S_0: M \rightarrow H, S_0(Tv) = v\,\implies\,S_0 \in B(M)$ according to closed mapping theorem. Define $S\,\vert_M = S_0, S\,\vert_{M^{\perp}} = 0\,\implies\,TS = P_M \in \mathcal{I}$. At last the proof claims that hence $\mathcal{I}$ contains the identity mapping. I am not sure if the last statement is correct.
Lemma 5.8: $H$ is a $\infty - dim$ Hilbert space. $T$ a compact operator iff $ran(T)$ contains NO closed $\infty - dim$ subspace.
In the original text there is only $\implies$ direction but the converse is also true. Let $H_{\leq 1}$ be the closed unit ball of $H$ and hence $\overline{TH_{\leq 1}}$ is a finite dimensional closed and bounded subspace and hence compact (with respect to the original topology).
Suppose $\mathcal{I}$ contains a projection $P$ onto an infinite-dimensional closed subspace $M$. In particular, $\dim M=\dim H=\aleph_0$ since $H$ is separable, so we can pick a Hilbert space isomorphism $U:M\to H$. Extend $U$ to an operator $T:H\to H$ which vanishes on $M^\perp$, and let $S:H\to H$ be the composition of $U^{-1}:H\to M$ with the inclusion $M\to H$. Then $TPS=1$, so $1\in\mathcal{I}$ and the ideal is not proper.
This argument does not use the assumption that $\mathcal{I}$ is closed, but it does use the assumption that $H$ is separable and $\mathcal{I}$ is two-sided. If $H$ has uncountable dimension, then the set of operators whose range is separable forms a proper closed two-sided ideal and contains projections onto infinite-dimensional subspaces. If $\mathcal{I}$ is only a one-sided ideal, then it could be the ideal of all operators that vanish on $M^\perp$ (for a left ideal) or operators whose range is contained in $M$ (for a right ideal).