 Solution

There is a cofinal branch. We give a proof that does not require distinct limit-level nodes to have different predecessor chains.

The set $S=\{\alpha <\lambda :\mathrm{cf}(\alpha )=\kappa \}$ is stationary. Indeed a strictly increasing continuous $\kappa$-sequence in any club set has supremum in that club set, below $\lambda$, of cofinality $\kappa$. At each $\alpha \in S$, there are fewer than $\kappa$ pairs of nodes on $T_{\alpha }$. For every pair whose predecessor chains below $\alpha$ differ, choose a height where they differ. Since $\mathrm{cf}(\alpha )=\kappa$, all these heights are bounded by some ${\beta }_{\alpha }<\alpha$. Consequently nodes in $T_{\alpha }$ having the same predecessor at ${\beta }_{\alpha }$ have identical predecessor chains below $\alpha$.

The Fodor lemma states that a regressive function on a stationary subset of a regular uncountable cardinal is constant on a stationary subset. Applying it here, ${\beta }_{\alpha }$ has a constant value $\beta$ on a stationary subset $S^{\prime }$. Choose $t_{\alpha }\in T_{\alpha }$ for each $\alpha \in S^{\prime }$. The level $T_{\beta }$ has fewer than $\kappa <\lambda$ nodes. Partitioning $S^{\prime }$ according to the predecessor of $t_{\alpha }$ at height $\beta$, one fiber $S^{\prime \prime }$ is stationary, since the union of fewer than $\lambda$ nonstationary sets is nonstationary. Let its common predecessor be $t$.

For $\alpha <{\alpha }^{\prime }$ in $S^{\prime \prime }$, the predecessor of $t_{{\alpha }^{\prime }}$ at level $\alpha$ and $t_{\alpha }$ have the same predecessor $t$ at level $\beta$. Their chains below $\alpha$ therefore agree. For each $\nu <\lambda$, choose $\alpha \in S^{\prime \prime }$ above $\nu$ and let $b_{\nu }$ be the predecessor of $t_{\alpha }$ at level $\nu$. The preceding comparison makes $b_{\nu }$ independent of that choice. The nodes $b_{\nu }$ form a chain through every level:This proves the uniformly narrow regular-height tree branch theorem. The uniform bound below the smaller regular $\kappa$ is stronger than merely bounding each level below $\lambda$.