I am interested in topological spaces such that whenever the space embeds into the <a href="http://en.wikipedia.org/wiki/Hilbert_cube">Hilbert cube</a>, the image of the embedding has a path-connected complement.

Any finite dimensional space has this property by an argument based on Alexander duality in a finite dimensional approximation of the Hilbert cube, see e.g. Lemma 2.1 in <a href="http://www.ams.org/journals/proc/1974-043-02/S0002-9939-1974-0334221-8/">"Characterization of finite-dimensional 𝑍-sets"</a>
by Kroonenberg [Proc. Amer. Math. Soc. 43 (1974), 421-427].

Are there other examples? 

UPDATE: 

1. First of all, a correction: the above mentioned Lemma 2.1 needs an assumption that the image of the embedding is closed. Without this assumption I can only show that the complement of a finite dimensional subset of a Hilbert cube is connected. (The proof is the same where instead of Alexander duality one uses the result that codimension two subspaces of $\mathbb R^n$ have connected complements).

2. The simplest example of a subset of the Hilbert cube with path-connected complement is that of deficiency $\ge 2$, where <i>deficiency of a subset</> is the number of coordinate projections mapping the subset to a point. I could not find any study of subsets of deficiency $\ge 2$. Rather, people studied closed subsets of infinite deficiency, or their images under ambient homeomorphisms which are also known as the $Z$-sets. So one could ask for conditions on a space
such that any embedding in the Hilbert cube is a $Z$-set. The definitive answer is given be Lenaburg in <a href="http://matwbn.icm.edu.pl/ksiazki/fm/fm85/fm8515.pdf">"Absolute $Z$-sets"</a>: any space with this property is countable!
Thus we reach a sad conclusion: looking at closed subsets of infinite deficiency does not give any new examples of spaces that always separate the Hilbert cube no matter how we embed them.