Take nonzero $x_0\in H$. Then
$$
H_0=\overline{\operatorname{span}}\{A^n x_0:\ n\in\mathbb N\cup\{0\}\}
$$
is a separable invariant subspace for $A$. If $H_0\subsetneq H$, take $x_1\in H_0^\perp$ and repeat. Using Zorn's Lemma (details at the end) one can construct a maximal family $\{x_j\}$ of such elements, and so $$H=\bigoplus_{j\in J}H_j,$$ with $H_j$ separable and invariant. If $J$ is infinite, split it into two disjoint equally infinite subsets $J_0$ and $J_1$, and now $\bigoplus_{J_0}H_j$ is a large invariant subspace with large orthogonal complement.
So now it remains to treat the case where $J$ is finite. We may as well assume $H$ is separable infinite dimensional. If there exists $M$ such that $P=E(M)$ and $I-P$ are infinite-dimensional, then $PH$ is invariant. Because it's infinite-dimensional with infinite-dimensional orthogonal complement in a separable space (so both have equal dimension), both are large.
We are now left with the case where either $E(M)$ or $I-E(M)$ is finite for all Borel sets $M$. Suppose there exist infinitely many pairwise disjoint Borel subsets with $E(M_k)$ finite. Then we can consider the invariant subspaces $H_k=E(M_k)H$ and again play the game of dividing them into two infinite families to get a large invariant subspace with large orthogonal complement. This leaves us with the case where any family of pairwise disjoint Borel subsets with $E(M_k)$ finite, is finite. A maximal family of such sets will be finite, then, and by taking the union we get $M$ such that $E(M)$ is finite and $E(N)$ is infinite for all $N\subset M^c$. If $E(M^c)\neq0$ and $M^c$ has at least two points we can get $E(N_1),E(N_2)$ infinite for $N_1\cup N_2=M^c$ and disjoint, and again we get $E(N_1)H$ invariant, large, with large orthogonal complement. This leaves us with the case where $E(M)$ is finite and $M^c$ has at most one point. Writing $P=E(M)$ we have $AP$ finite rank and $$A=AP+\lambda(I-P)=\lambda I+(A-\lambda I)P$$ as Halmos says. Since $P$ is finite rank, $I-P$ is infinite. Writing $I-P=Q_1+Q_2$ pairwise orthogonal and infinite, the subspaces $Q_1H$ and $Q_2H$ are invariant, countably infinite dimensional, and orthogonal. So $Q_1H$ is invariant, large, with large orthogonal complement.
Use of Zorn's Lemma. Let
$$
\mathcal F=\{(x_j):\ \langle A^nx_j,A^mx_k\rangle=0\ \text{ for all } k\ne j,\ \text{ and all } n,m\},
$$
ordered by inclusion. This is trivially nonempty since $A$ is selfadjoint and so the orthogonal complement of an invariant subspace is invariant. Given a chain $\{(x_j)_{j\in J_\alpha}: \alpha\}$, where $J_\alpha\subset J_\beta$ if $\alpha\leq\beta$, let $J=\bigcup_\alpha J_\alpha$. Then $(x_j)_{j\in J_\alpha}\subset (x_j)_{j\in J}\in\mathcal F$ for all $\alpha$, so it is an upper bound. By Zorn's Lemma there exists a maximal element $(x_j)_{j\in J}$. With $H_j=\overline{\operatorname{span}}\{A^nx_j:\ n\}$, let $\tilde H=\bigoplus_jH_j$. Then $\tilde H=H$, because otherwise there exists $y\in \tilde H^\perp$, and since $\tilde H^\perp$ is invariant, this contradicts the maximality of $(x_j)$. So
$$
H=\bigoplus_jH_j.
$$
Best Answer
"Copy of X" here means "space that is isomorphic to X". So the condition is indeed that $H$ should contain infinitely many subspaces, each of which is isomorphic to the orthogonal complement $H^\perp$, and which are pairwise orthogonal to each other.
A concrete example might be $K = L^2(\mathbb{R})$, $H = L^2(0,\infty)$. Then $H^\perp = L^2(-\infty,0)$. Let $H_i = L^2(i, i+1)$, which you can show is isomorphic to $L^2(-\infty, 0)$, and the $H_i$ are all pairwise orthogonal. So $H$ is an example of a large subspace.
The notation $K \setminus H$ here doesn't mean set difference, but orthogonal complement. So $K \setminus H$ is just the orthogonal complement of $H$ in $K$, which you could also denote $H^\perp$ when the ambient space $K$ is understood. It is certainly a closed subspace of $K$, and it makes sense to speak of its dimension.
The $K \setminus H$ notation is fairly common, and although it is technically ambiguous with set difference, in practice it is usually clear from context which one is meant.
Let $H$ be a subspace of the separable Hilbert space $K$. It is clear that a large subspace must be infinite-dimensional (it contains infinitely many nontrivial orthogonal subspaces).
For the converse, suppose $H$ is infinite-dimensional, and fix an orthonormal basis $\{e_1, e_2, \dots\}$ for $H$. Let $H^\perp$ be its orthogonal complement.
Suppose first that $H^\perp$ has finite dimension $n$. For each $i > 0$ let $H_i$ be the subspace of $H$ spanned by $\{e_{ni+1}, \dots, e_{ni+n}\}$. Each $H_i$ is isomorphic to $H^\perp$ because they have the same dimension $n$, and the $H_i$ are pairwise orthogonal because they are spanned by disjoint subsets of an orthogonal set. Thus $H$ is large.
Otherwise, suppose $H^\perp$ is infinite-dimensional; since it is separable, it has a countable orthonormal basis $\{f_1, f_2, \dots\}$. Let $\{ k_{1,1}, k_{1,2}, \dots\}$, $\{k_{2,1}, k_{2,2}, \dots\}$ be infinitely many disjoint sequences in $\mathbb{N}$; for instance you could enumerate the primes as $p_n$ and let $k_{n,i} = p_n^i$. Let $H_i$ be the closed linear span of $\{e_{k_{i,1}}, e_{k_{i,2}}, \dots\}$. Each $H_i$ is isomorphic to $H^\perp$, because $H^\perp, H_i$ are both infinite-dimensional separable Hilbert spaces, or because they both have orthonormal bases of the same cardinality $\aleph_0$. And the $H_i$ are again pairwise orthogonal because they are the closed linear spans of disjoint orthogonal sets. Thus $H$ is also large in this case.
In essence, it's really just the fact that a countably infinite set can be written as a countable union of countable sets (or of finite sets), i.e. that $\aleph_0 = \aleph_0^2$, applied to dimension instead of cardinality.
The term "large subspace" seems to have been introduced by Halmos and I haven't seen it used by anyone else; the concept is not one that I think comes up especially often. I wouldn't particularly expect there to be a lot of reference material on the topic. It does appear in Halmos's Hilbert Space Problem Book on page 131, where he says that "the idea has appeared before, even if the term has not".