In my previous question it was established that if $X$ is a metrizable, connected, locally path connected space and $K\subset X$ is compact, then there is a Peano continuum $L\subset X$ such that $K\subset L$. This motivated the following questions.
Let $X$ be as above and $\pi_k(X)$ is trivial, for $k=0,...,n$ (resp $X$ is contractible). If $K\subset X$ is a Peano continuum, can we find a compact $L\subset X$ such that $K\subset L$ and $\pi_k(L)$ is trivial, for $k=0,...,n$ (resp $L$ is contractible)?
Regarding the "contractible version", here is an idea that does not work: let $F:X\times[0,1] \to X$ be a homotopy from the identity to a constant. It is tempting to try to show that $F(K\times [0,1])$ is the set that we are looking for. However, if we started with $K$ a singleton, the obtained set can be any Peano continuum, and so not necessarily contractible.
Edit: There is a good candidate for a counterexample suggested by HJRW. Let $X$ be the Whitehead manifold, which is a contractible $3$-dimensional manifold, not homeomorphic to $\mathbb{R}^3$. It is proven in the paper Brick & Mihalik - The QSF property for groups and spaces, that $X$ fails the so called QSF property. This amounts to existence of subcomplex $A$ of $X$ such that there is no celluar map $f:K\to X$ of a finite simply connected complex such that $f|_{f^{-1}(A)}\to A$ is a homeomorphism. In particular, there is no finite simply connected subcomplex $L\supset A$.
Hence, we are almost done: there is a compact set (in a contractible manifold) that cannot be included into a simply-connected subcomplex. The missing ingredient therefore is:
Can every simply connected $M\subset X$ be included in a simply connected finite subcomplex of $X$?
Interestingly, in Definition 1.5 this article, the QSF property is defined with no reference to complexes, and in fact is exactly what i need. However, that's the only place i saw this definition, so it's not 100% clear it is equivalent to the more common one.
