Slim Kurepa tree at a singular strong limit cardinal of uncountable cofinality For a strong limit cardinal $\kappa$ the notion of $\kappa$-Kurepa tree is trivial: the full binary tree is a $\kappa$-Kurepa tree.  Accordingly, we consider the following strengthening:
A slim $\kappa$-Kurepa tree is a tree $T$ of height $\kappa$ such that for every infinite $\alpha < \kappa$ the $\alpha$-th level of $T$ has cardinality $\left| \alpha \right|$, and $T$ has more than $\kappa$ many branches.
If $\kappa$ is a strong limit cardinal of countable cofinality, it's easy to construct a slim $\kappa$-Kurepa tree.  On the other hand, if $\kappa$ is measurable (or just ineffable) then there is no slim $\kappa$-Kurepa tree.  If $\kappa$ is inaccessible, then my understanding from comments here is that there is a $\mathord{<}\kappa$-closed forcing to create a slim $\kappa$-Kurepa tree (but this destroys measurability.)  What about the uncountable cofinality singular case?

If $\kappa$ is a singular strong limit cardinal of uncountable cofinality, can there exist a slim $\kappa$-Kurepa tree?

EDIT: This turned out to be fairly easy; see my answer below.  However, I would like to know where I can find this result proved (or at least mentioned) in print.  So I will accept the first answer that tells me this.
 A: The answer is no.  Martin Zeman showed me this proof.  (Any mistakes were probably introduced by me.)
Let $\kappa$ be a singular strong limit cardinal of uncountable cofinality and let $T$ be a tree of height $\kappa$ such that for every $\alpha < \kappa$ the $\alpha$-th level of $T$ has cardinality $\left|\alpha \right|$.
We will show that $T$ has at most $\kappa$ many cofinal branches.
Let $\gamma$ be the cofinality of $\kappa$ and
let $(\kappa_\xi : \xi < \gamma)$ be a continuous increasing sequence of cardinals that is cofinal in $\kappa$.  For every $\xi < \gamma$ let $(b^\xi_\alpha : \alpha < \kappa_\xi)$ enumerate the $\kappa_\xi$-th level of $T$.
For every branch $b$ of $T$, by a pressing-down argument there is a stationary subset $S \subset \gamma$ and an ordinal $\beta < \kappa$ such that for every ordinal $\xi \in S$ we have $b \restriction \kappa_\xi = b^\xi_\alpha$ for some $\alpha < \beta$.
So every branch is determined by a stationary subset $S \subset \gamma$ and a bounded function $S \to \kappa$, and there are only $\kappa$ many such functions.
A: The following is proved by Erdos-Hajnal-Milner in ``On sets of almost disjoint subsets of a set. Acta Math. Acad. Sci. Hungar 19 1968 209–218'', from which the required result follows
Theorem. ssume $\aleph_0 < cf(\kappa) < \kappa$ and $\forall \theta< \kappa, \theta^{cf(\kappa)} < \kappa.$ Let $F \subseteq P(\kappa)$
be such that $\{\alpha < \kappa: |F \restriction \alpha| \leq \alpha   \}$
is stationary. Then $|F| \leq \kappa.$
