On Lipschitz embeddability of certain compact metric spaces into $\mathbb{R}^n$ Suppose that $K$ is a compact metric space which is homeomorphic to a subspace of $\mathbb{R}^n$. Does there exist $f:K\rightarrow \mathbb{R}^n$ which is one-to-one and Lipschitz? 
 A: No. There is a length metric counter-example in dimension 3. See Theorem 1 in this paper.
Let me briefly explain the construction here. For a large $n$, consider a unit segment $[p_nq_n]\subset \mathbb R^3$ and let $U_n$ be its neighborhood of radius $2/n$. Inside $U_n$, consider a "chain" formed by $n$ small solid tori. Each solid torus is a $(1/10n)$-neighborhood of a circle of radius $1/n$, and each of these circles is linked to the next one. The first link in the chain is nearby $p_n$ and the last one is nearby $q_n$. In each of these solid tori introduce a conformal Riemannian metric as follows. At the middle circle, the conformal factor is very small, so that the length of this circle equals $1/n^2$. And the conformal factor equals 100 almost everywhere between the middle circle and the boundary of the torus. So despite the fact that the middle circle is very short, the distances between points outside the torus are no shorter than the Euclidean distances. It follows that the distance between $p_n$ and $q_n$ in the resulting Riemannian metric is at least 1.
Repeat this constriction for $n=10,100,1000,\dots$, choosing the segments $[p_nq_n]$ so that they converge to some $[pq]$ (but the neighborhoods $U_n$ are disjoint). This yields a length metric on $\mathbb R^3$ which defines the standard topology (i.e., the identity is a homeomorphism). Restrict it to a compact ball containing all the segments of the construction.
There is no Lipschitz homeomorphism from this ball with this metric to $\mathbb R^3$. Indeed, suppose that $f$ is a 1-Lipschitz homeomorphism. Consider the image of the neighborhood of $[p_nq_n]$. The images of the chained circles have lengths (and hence diameters) at most $1/n^2$, Since they are linked, each next circle is within distance $1/n^2$ from the previous one. So the endpoints of the chain are mapped to points at distance at most $2/n$ or so from each other: $|f(p_n)-f(q_n)|\le 2/n$. It follows that $f(p)=f(q)$, a contradiction.
