Slim Kurepa tree at a singular strong limit cardinal of uncountable cofinality

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.

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 $|F| \leq \kappa.$