Recall that a plane continuum is a closed, bounded, connected subset of the plane. It is non-degenerate if it contains at least two points. (We may sometimes just say "continuum" even if we mean a non-degenerate one.)

Kevin Johnson asked if every plane continuum, which is not a line segment with slope $1$ or $-1$, must cross a dyadic square.
(By a line segment we mean a straight line segment. Links given at the end.)

The present contributor asked if there is a hereditarily square-boxed plane continuum which is not a line segment with slope $1$ or $-1$. The question below is related.

Definition.
Let $C$ be a closed, bounded subset of the plane. A bounding box for $C$ is a rectangle $B=[a,b]\times[c,d]$ that contains $C$ and such that no smaller rectangle with vertical and horizontal sides contains $C$.
(It follows that each side of $B$ intersects $C$.)
When $B$ happens to be a square then we call $C$ square-boxed.

Clearly if $J$ is a line segment with slope $1$ or $-1$ then every subcontinuum of $J$ is also such a line segment and its bounding box must be a square. (We say that $J$ is hereditarily square-boxed.)
On the other hand, if $D$ is the line segment connecting points with coordinates $(0,0)$ and $(2,1)$ then the bounding box for every subcontinuum of $D$ is a rectangle whose width $w$ is always twice as big as its height $h$. In particular, $D$ has no square-boxed non-degenerate subcontinuum.

Question.
If $C$ is a hereditarily indecomposable plane continuum, must $C$ have at least one non-degenerate square-boxed subcontinuum?

If so, this will provide a positive answer to Kevin Johnson's question in the class of hereditarily indecomposable continua (as we will comment further below). Recall that a continuum is indecomposable if it not the union of two proper subcontinua. It is hereditarily indecomposable if every subcontinuum is indecomposable. The pseudoarc is an example of a hereditarily indecomposable plane continuum. It is known that, generically, most plane continua are hereditarily indecomposable (even if this property may appear pathological).

Here are some related comments and examples.

Definition.
Let $B=[a,b]\times[c,d]$ be a rectangle in the plane, with $w=b-a$ and $h=d-c$ being its width and height. Define slenderness of $B$ as $\sigma(B)=\frac{h}w$. Clearly $0<\sigma(B)<\infty$ whenever $w$ and $h$ are positive, and if $S$ is a square (so $w=h$) then $\sigma(S)=1$. But we would like to also consider "rectangles" with either zero width (a vertical line segment) or zero height (a horizontal line segment) and in this case $0\le\sigma(B)\le\infty$. To avoid working with $\infty$ we adopt a different but closely related definition, with values in the interval $[-1,1].$ We call $\lambda(B)=\frac{-w^2+h^2}{w^2+h^2}$ the lankiness, or lank rank of $B.$ A vertical line segment (when $w=0<h$) has lankiness $1$ (it is $100\%$ lanky), while a horizontal line segment (when $w>0=h$) has lankiness $-1$, and a square has lankiness $0$. (The lankiness of a singleton remains undefined, unless we want to allow the entire interval $[-1,1]$ as its value).
If $C$ is a closed, bounded subset of the plane then we also define its slenderness $\sigma(C)=\sigma(B)$ and its lankiness $\lambda(C)=\lambda(B)$ where $B$ is the bounding box for $C.$

It is easily seen that if $C$ is a circle and $\lambda$ is any number in $(-1,1)$ then there is a subcontinuum $K$ of $C$ with $\lambda(K)=\lambda$. The same holds more generally if $C$ is any continuously differentiable Jordan curve in the plane (where a Jordan curve is a closed plane curve, i.e. a loop, with no self-intersections).

On the other hand there are Jordan curves $J$, as shown in the figure below, such that not every $\lambda\in(-1,1)$ is the lank rank of some subcontinuum of $J$. Figure 1.(a) shows a Jordan curve $J_a$ consisting of some line segments of slope $\pm2$ and $\pm3$. It has a square-boxed subcontinuum $S$ (with $\sigma(S)=1$ and $\lambda(S)=0$) and another subcontinuum $G$ with $\sigma(G)=3$ and $\lambda(G)=\frac{-1^2+3^2}{1^2+3^2}=\frac45$, but no subcontinuum $K$ with either $\sigma(K)<1$ or $\sigma(K)>3$. (That is, no $K$ with either $\lambda(K)<0$ or $\lambda(K)>\frac45$.)

For any plane continuum $C$ let $\Lambda(C)$ be the set of all $\lambda(K)$ where $K$ runs through all non-degenerate subcontinua of $C.$ We will argue later that $\Lambda(C)$ must always be convex, so it is either a sigleton or an interval (including or not some of its endpoints).

For example if $J$ is a line segment with slope $1$ or $-1$ then $\Lambda(J)=\{0\}$. If $D$ is the line segment connecting points $(0,0)$ and $(2,1)$ then $\Lambda(D)=\{\frac{-3}5\}$. If $C$ is any continuum with non-empty interior then $\Lambda(C)=[-1,1].$ If $C$ is a circle then $\Lambda(C)=(-1,1).$ If $R$ consists of the edges of some rectangle (even when the interior is not included) then $\Lambda(R)=[-1,1].$

If $J_a$ is the continuum on Fig.1(a) (having some line segments of slope $\pm2$ and $\pm3$) then $\Lambda(J_a)=[0,\frac45]$. If we stretch this continuum vertically by
a factor of $\frac32$ then we obtain the continuum $J_b$ on Fig.1(b) (having some line segments of slope $\pm3$ and $\pm\frac92$), with $\Lambda(J_b)=[\frac5{13},\frac{77}{85}]$ (take these numbers for granted, or verify them if you wish). In particular, $J_b$ has no square-boxed subcontinuum. If we add some line segments of slope $\pm\frac13$ then we obtain a continuum $J_c$ in Fig.1(c) with $\Lambda(J_c)=[\frac{-4}5,\frac45]$ (and in particular $J_c$ has a square-boxed subcontinuum).

A pseudoarc could be constucted as the limit of a sequence of curves that ziz-zag a lot and keep visiting back near, and going away from, points where they had already been. (Take this as the present contrubutor's limited understanding of the pseudoarc.) The point being that, being so complicated, the pseudoarc ought to have subcontinua of all possible lank ranks (except $\pm1$ since a preudoarc contains no path connected subcontinua, and in particular no vertical or horizontal line segments). Since we do not have a proof for the "ought ot have" part, we pose it as a question.

Question.
Let $C$ be any hereditarily indecomposable plane continuum.
(a) must $C$ have at least one non-degenerate square-boxed subcontinuum?
(b) must it be the case that $\Lambda(C)=(-1,1)$ ?

Although I do not know the answer to either (a) of (b), it is easy to see that parts (a) and (b) of the above question are equivalent. That (b) implies (a) is clear (as $K$ must be square-boxed if $\lambda(K)=0$), so we will indicate why (a) implies (b). Assume (a) and start with any hereditarily indecomposable plane continuum $C$, and then stretch it, either vertically or horizontally, to obtain a homeomorphic hereditarily indecomposable plane continuum $T$. By assumption (a), $T$ must have a square-boxed subcontinuum $L$. Now compress $T$ back onto $C$ (undoing exactly whatever stretching was done), the result being that $L$ is transformed onto a subcontinuum $K$ of $C$, and we could make $\lambda(K)$ to be any number $\lambda\in(-1,1)$, as we wish, by adjusting the amount (and direction) of stretching that was done.

Here is a sketch of the argument that $\Lambda(C)$ must always be convex (for any non-degenerate continuum $C$). Suppose $\lambda_1,\lambda_2\in\Lambda(C)$ with $\lambda_1<\lambda_2$ and pick any
$\lambda$ with $\lambda_1<\lambda<\lambda_2$. We need to show that there is a subcontinuum $K$ of $C$ with $\lambda(K)=\lambda$. Pick non-degenerate subcontinua $K_1,K_2$ of $C$ with $\lambda(K_1)=\lambda_1$ and $\lambda(K_2)=\lambda_2$. We can gradually and continuously "grow" $K_1$ till it coincides with $C$. (Main idea, take a rectangle slightly larger than the bounding box of $K_1$, and with edges that do not intersect $K_1$, then apply the boundary bumping theorem.) More precisely, there is an increasing by set inclusion chain of continua (note that "chain" here is used with its set-theoretic meaning, as a chain of nested continua; and not with its often encountered continuum theoretic meaning involving "adjacent links" that intersect)... so there is an increasing by set inclusion chain of continua (or a nested family of continua) starting at $K_1$ and ending at $C$ such that the width $w$ as well as the height $h$ of the respective bounding boxes are continuous non-decreasing functions. (In particular every continuum in this chain is non-degenerate, being bigger than $K_1$, so lank rank is always well-defined.)
Similarly there is another increasing by set inclusion chain of continua, starting at $K_2$ and ending at $C$. Traverse the first chain from $K_1$ to $C$, and then follow by traversing in reverse the second chain from $C$ to $K_2$, then the lank rank of the contunia changes from $\lambda(K_1)=\lambda_1$ to $\lambda(C)$, and from there to $\lambda(K_2)=\lambda_2$. So altogether the lank rank changes continuously from $\lambda_1$ to $\lambda_2$, hence by the intermediate value theorem there is some moment of time (i.e. a continuum $K$ in at least one of the two chains) such that $\lambda(K)=\lambda.$

Here is a link to Kevin Johnson's question about a continuum crossing dyadic squares:
URL (version: 2013-01-31):
Does every connected set that is not a line segment cross some dyadic square?

Here is a link to a related question that I posted a few days before I posted the present one.
URL (version: 2023-05-03):
Are there hereditarily square-boxed plane continua?

It is indicated in the latter question that, unless a continuum if hereditarily square-boxed, then it must cross some dyadic square.

So, if every hereditarily indecomposable plane continuum $C$ must contain at least one square-boxed continuum (which is the question asked here), then $\Lambda(C)=(-1,1)$, so $C$ cannot be hereditarily square-boxed, so it much cross some dyadic square.

(Disclaimer. I seem to promote above the idea that $\Lambda(C)=(-1,1)$ for every hereditarily indecomposable plane continuum $C$, but I feel quite lost with this question, and I might be imagining things due to my incomplete understanding. I am curious.)