text: mg.metric geometry - On Lipschitz embeddability of certain compact metric spaces into $\mathbb{R}^n$ - MathOverflow
  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? mg.metric-geometry share|cite|improve this question asked Mar 22 '13 at 15:10
  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.
  share|cite|improve this answer answered Mar 22 '13 at 16:49
  Thank you for your answer. The question remains open for n=1,2. It's good to keep in mind that when $K$ is a compact and countable metric space, there's always a Lipschitz injection from $K$ into $\mathbb{R}$. Indeed, for each distinct $x,y\in K$, $H_{x,y} = \{f\in Lip(K):f(x)=f(y)\}$ is a hyperplane of the Banach space $Lip(K)$ of real-valued Lipschitz functions on $K$. Baire's Theorem implies that $Lip(K)\neq \cup_{x\neq y} H_{x,y}$, thus there exists $f\in Lip(K) \setminus \cup_{x\neq y}H_{x,y}$ -- which has to be one-to-one. (this result is from a personal communication from G. Godefroy)
  I think it is true for $n=1$ and false for $n=2$. For $n=2$ you can do a similar construction using circles and points inside them (but the resulting metric is no longer intrinsic). For $n=2$ and an intrinsic metric on a disc, it is an open problem although not a very popular one. For $n=1$, it seems that one can can construct a Lipschitz embedding using the linear order on the real line: divide the interval containing the set in half, then in 4 pieces, etc, and choose images of the division points carefully. – Sergei Ivanov