A continuum which is both Suslinean and non-Suslinean? Continuum means compact connected metrizable with more than one point.
A continuum is  Suslinean if every collection of pairwise disjoint subcontinua is countable.
There is an apparent contradiction in the literature I would like to resolve...
In Example 3 at the end of Paper A, there is constructed a continuum $Y:=X/\sim$ which is the quotient of another continuum $X$. It is clear from the construction that $X$ is Suslinean, and since the quotient map is monotone  $Y$ is also Suslinean. Actually, the author claims $X$ and $Y$ are hereditarily locally connected, and such continua are automatically known to be Suslinean.
On the other hand, $Y$  is the closure of a ray (a one-to-one continuous image of $[0,1)$) such that the ray is first category in $Y$.  In other words, the ray and its complement are each dense in $Y$.  It follows that there is a sequence of pairwise disjoint arcs in $Y$ which converges to $Y$ in the Hausdorff distance.  By Theorem 30 in Paper B, $Y$ is non-Suslinean.
Question 1: Am I correct in finding a contradiction?
Question 2: Is Example 3 correct? Specifically, why is it Hausdorff (and therefore metrizable)?  I found a couple of typos, e.g. $A_n$ should be $C_n\cup \bigcup ...$ and $\overline{z_1 z_1}$ should be $\overline{z_1 z_2}$, but otherwise it seems okay.
If there is a problem with Theorem 30, then I think the error must be in the proof of Lemma 29.  Specifically there is a claim that $H_W$ is non-degenerate since it is an inverse limit of non-degenerate continua. This is a false statement in general because we can take the continua $[0,1/n]$ with bonding maps the inclusions $[0,1/(n+1)] \hookrightarrow[0,1/n]$, and the inverse limit is just the single point $\langle 0,0,..\rangle$. However, since each factor of the inverse limit for $H_W$ projects into $\partial U$ and $\partial V$ in the first factor, the inverse limits on preimages of these boundary sets should be non-empty and disjoint subsets of $H_W$...
Upon closer inspection, the proof of Lemma 29 (Paper B) is flawed in more ways than one.  I'm not sure it can be saved.
At this point I would lean toward the example being correct.  I really just need to see that $Y=X/\sim$ is Hausdorff.

Paper A:
Tymchatyn, E. D., Some rational continua, Rocky Mt. J. Math. 13, 309-319 (1983). ZBL0514.54022.
Paper B:
Mouron, Christopher, The topology of continua that are approximated by disjoint subcontinua, Topology Appl. 156, No. 3, 558-576 (2009). ZBL1165.54012.
 A: Example 3 in Paper A is indeed a counterexample to Theorem 30 in Paper B.  
Lemma 29 must also be false because it implies Theorem 30.
The only thing left to verify is that $Y=X/\sim$ (from Example 3) is Hausdorff, so that $Y$ is actually a metrizable continuum.  This can be proved in a few steps:


*

*For every compact $K\subseteq [0,1]$ the set $$\widehat{K}:=K\cup \bigcup \big\{\overline{z_1 z_2}:z_1\text{ and }z_2\text{ are consecutive endpoints of some }C_n \text{ and }(z_1\in K\text{ or }z_2\in K)\big\}$$ is a closed (compact) subset of $X$.

*If $U$ is any open subset of $X$ containing an element $y\in Y$, then   $$V=X\setminus\widehat{[0,1]\setminus U}$$ is an open subset of $X$ which is a union of members of $Y$, and $y\subseteq V\subseteq U$. 

*$Y$ is Hausdorff by  2 and normality of $X$.


We can argue  1 as follows.  Let $x$ be any point in the closure of $\hat K$.  Suppose for a contradiction that  $x\notin \hat K$. Then $x\in [0,1]\setminus K$ and there is a sequence of semi-circular arcs $\overline{z^n_1 z^n_2}\in Y$ with $z^n_1\in K$ and $z^n_2\to x$. (I assume without loss of generality that the subscripts on the $z$'s are arranged in this way). 
Since $d(x,K)>0$ this means infinitely many semicircles have radius greater than some positive constant.  But this is impossible by the construction of $X$. Therefore $x\in \hat K$   and $\hat K$ is compact.
Now 1 $\Rightarrow$ 2 $\Rightarrow$ 3 and the proof is complete.
