Is every metric continuum almost path-connected? The question was motivated by this question of Anton Petrunin.
By a metric continuum we understand a connected compact metric space.
Let $p$ be a positive real number. A metric continuum $X$ is called $\ell_p$-almost path-connected if for any points $x,y\in X$ and any $\varepsilon>0$ here exists a family $\big((a_n,b_n)\big)_{n\in\omega}$ of pairwise disjoint open intervals in the unit segment $[0,1]$ and a continuous map $\gamma:[0,1]\setminus\bigcup_{n\in\omega}(a_n,b_n)\to X$ such that $\gamma(0)=x$, $\gamma(1)=y$ and $\sum_{n=0}^\infty d_X(\gamma(a_n),\gamma(b_n))^p<\varepsilon$.
It is easy to see that each almost $\ell_p$-connected metric continuum is $\ell_q$-almost connected for any $q\ge p$.
By my answer to the question of Anton Petrunin, each plane continuum is almost $\ell_1$-connected. By analogy it can be shown that each continuum in $\mathbb R^3$ is $\ell_2$-connected.
Problem. Is there a metric continuum which is not almost $\ell_1$-path connected? not almost $\ell_p$-connected for every $p<\infty$?    
 A: Yes there are such examples.
Assume the sequence $\varepsilon_n$ is very fast converging to $0$.
Consider a sequence of short $\varepsilon_n$-crooked maps between intervals $\mathbb{J}_n\to \mathbb{J}_{n-1}$. 
Its inverse limit is a pseudoarc $\mathbb{J}_\infty$; denote by $\phi_n\colon\mathbb{J}_\infty\to \mathbb{J}_n$ the projections.
Equip $\mathbb{J}_\infty$ with the maximal metric such that 
$$|x-y|_{\mathbb{J}_\infty}\le \tfrac1{2^{n/p}}+|\phi_n(x)-\phi_n(y)|_{\mathbb{J}_n}.$$
Assume that $\mathbb{J}_\infty$ is $\ell_p$-almost path-connected.
Let $\gamma$ be the path that connects the ends of $\mathbb{J}_\infty$.
It can not have more then one jump of size $1$; the jump brakes $\alpha$ in two arcs one of which has diameter at least $1-\varepsilon_1$;
this arc can not have more than 2 jumps of length $\tfrac12$, so one of the subarcs has diameter at least  $1-\varepsilon_1-2\cdot \varepsilon_2$ and so on. At the end of the day you see that $\mathbb{J}_\infty$ contains a subarc of $\alpha$ of positive diameter. But $\mathbb{J}_\infty$ has no arcs, a contradiction.
The given construction is nearly identical to  Example 4.2 in my paper on intrinsic isometries.
