The answer to this question is positive. A required path $\gamma$ can be constructed inductively using the following

**Lemma.** For any continuum $P\subset\mathbb R^2$, distinct points $x,y\in P$, and $\varepsilon>0$ there are continua $P_1,\dots,P_{k}\subset P$ of diameter $<\varepsilon$, and horizontal or vertical arcs $A_0,\dots,A_{k+1}$ in the plane such that

1) $\{x\}=A_0$, $\{y\}=A_{k+1}$;

2) $\sum_{i=1}^k\lambda(A_i\setminus P)<\varepsilon$;

3) for every $i\in\{1,\dots,k\}$ the continuum $P_{i}$ intersects the arcs $A_{i-1}$ and $A_i$.

*Proof.* Replacing $P$ by a smaller subcontinuum, we can assume that $P$ is irreducible between points $x$ and $y$, which means that any subcontinuum of $P$ that contains $x$ and $y$ coincides with $P$.

The irreducibility of $P$ implies that for any rational numbers $a<b$ the (closed) set $\{x\in \mathbb R:\{x\}\times[a,b]\subset P\}$ is nowhere dense in $\mathbb R$. Consequently the set $$G_1:=\{x\in\mathbb R:P\cap(\{x\}\times\mathbb R)\mbox{ is zero-dimensional}\}$$is dense $G_\delta$ in $\mathbb R$. Choose an increasing sequence $a_0,\dots,a_n\in G_1$ such that $P\subset [a_0,a_n]\times\mathbb R$ and $|a_i-a_{i-1}|<\frac12\varepsilon$ for all positive $i\le n$. For every $i\in\{0,\dots,n\}$ the inclusion $a_i\in G_1$ implies that the set $Z_i:=\{y\in \mathbb R:(a_i,y)\in P\}$ is zero-dimensional and hence nowhere dense in $\mathbb R$.

By analogy, the set $$G_2:=\{y\in\mathbb R:P\cap(\mathbb R\times \{y\})\mbox{ is zero-dimensional}\}$$is dense $G_\delta$ in $\mathbb R$. Then we can choose a monotone sequence $b_0,\dots,b_m\in\mathbb G_2\setminus\bigcup_{i=0}^nZ_i$ such that $P\subset \mathbb R\times [b_0,b_m]$ and $|b_i-b_{i-1}|<\frac12\varepsilon$ for all positive $i\le m$.

For every $(i,j)\in\{1,\dots,n\}\times\{1,\dots,m\}$ consider the cube $Q_{i,j}:=[a_{i-1},a_i]\times[b_{j-1},b_j]$ and its boundary $\partial Q_{i,j}$ in the plane. The choice of the points $a_{i-1},a_i,b_{j-1},b_j$ guarantees that the intersection $P\cap\partial Q_{i,j}$ is zero-dimensional and does not contain the vertices of the cube $Q_{i,j}$.

Let $V:=\bigcup_{i=0}^n\{a_i\}\times [b_0,b_m]$ and $H:=\bigcup_{j=0}^m[a_0,a_n]\times\{b_j\}$ be the unions of vertical and horizontal sides of the cubes $Q_{i,j}$. Let $\mathcal C$ be the family of connected components of the symmteric difference $\Xi:=(V\cup H)\setminus (V\cap H)$. Each component $C\in\mathcal C$ is a vertical or horizontal interval of length $<\frac12\varepsilon$. By the regularity of the Lebesque measure on the real line, there is a disjoint finite family $\mathcal I$ of compact connected subsets of $\Xi\cup\{x,y\}$ such that

$\bullet$ $\{x\},\{y\}\in\mathcal I$;

$\bullet$ $P\cap\Xi\subset \bigcup\mathcal I$;

$\bullet$ $\lambda(\bigcup\mathcal I\setminus P)<\varepsilon$;

$\bullet$ each set $I\in\mathcal I$ meets the set $P$.

Now consider the finite graph $\Gamma$ whose set of vertices coincides with $\mathcal I$ and two distinct intervals $I,J\in\mathcal J$ form an edge of the graph $\Gamma$ if there exists a square $Q_{i,j}$ and a connected component of $P\cap Q_{i,j}$ that intersects both arcs $I$ and $J$. It can be shown that the graph $\Gamma$ is connected, so we can find a sequence $\{x\}=A_0,A_1,\dots,A_{k+1}=\{y\}$ of pairwise distinct vertices of the graph $\Gamma$ such that for every $i\le k$ the unordered pair $\{A_i,A_{i+1}\}$ is an edge of $\Gamma$. The latter means that $A_{i-1}$ and $A_i$ are intersected by some subcontinuum $P_i$ of $P$ that is contained in some square $Q_{j,l}$ and hence has diameter $<\varepsilon$.

It is clear that the sequences $P_1,\dots,P_k$ and $A_0,\dots,A_{k+1}$ have the required properties.

**Remark.** Unfortunately, the proof of the Lemma essentially uses the planarity of the continuum $P$. It would be interesting to know if the problem of Anton Petrunin still has affirmative answer for continua in arbitrary (finite-dimensional) Banach spaces.