The following is Theorem $2.8$ in Chapter 13 of Hrbacek and Jech's "Introduction to Set Theory" ($3^{\text{rd}}$ edition):
Theorem $\bf{2.8}$ Let $\kappa$ be a measurable cardinal. If $T$ is a tree of height $\kappa$ such that each node has less than $\kappa$ immediate successors, then $T$ has a branch of length $\kappa$.
The proof begins with:
Proof. $\;$ For each $\alpha \lt \kappa$, let $T_{\alpha}$ be the set of all $s \in T$ of height $\alpha$. Since $\kappa$ is a strongly inaccessible cardinal, it follows, by induction on $\alpha$, that $\vert T_{\alpha}\vert \lt \kappa$ for all $\alpha \lt \kappa$.
I'm having trouble understanding why this is true for $T_{\alpha}$, where $\alpha$ is a limit ordinal. If, for every $x, y \in T_{\alpha}$, $\{z \in T \; \vert \; z \lt x\}$ = $\{z \in T \; \vert \; z \lt y\}$ implies $x = y$, then I've been able to bound the size of $T_{\alpha}$, given that $\kappa$ is strongly inaccessible. But how does one bound $\vert T_{\alpha}\vert$ when each set of predecessors can correspond to multiple elements of $T_{\alpha}$?
The problem is of course at limit stages. But any point at the limit levels is determined by the branch below it. So we just need to understand how big can a tree of height $\alpha$ be, if all the levels are smaller than $\kappa$.
Since $\kappa$ is regular, we can find an upper bound on the sizes of the levels, say $\lambda$. So, now the tree has size $\alpha\cdot\lambda$. And how many branches can it have? At most $2^{\alpha\cdot\lambda}$, which is less than $\kappa$.