Let T be a tree such that every leaf is adjacent to a vertex of degree at least 3. Show that there are two leaves with a common neighbor.

Suppose tree is finite.

If no two leafs have common neighbour, then when deleting all leafs we get a graph $H$ with all vertices of degree $2$ or more in $H$. So $H$ has a cycle. But then $T$ has also a cycle. A contradiction.

You could also root you tree anywhere and take a leaf of maximal depth $\delta$. Since its parent (of depth $\delta - 1$) is of degree 3, there is another node of depth $\delta$, and it is a leaf (because there is no node of depth $\delta + 1$).