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A 3-connected graph contains a cycle, and any cycle is a complete graph subdivision $T{K}_3$.

For $T{K}_4$, choose a cycle $C$ for which some $C$-bridge has at least three attachment vertices $a,b,c$; such a choice is always possible in a 3-connected graph. Indeed, every component outside a cycle has at least three attachments, since one or two would form a vertex cut, and if a chosen cycle is spanning, a chord can be used to choose a smaller cycle and expose the remaining bridge. In a minimal connected part of that bridge joining $a,b,c$, suppress degree-two vertices. The result has a branch vertex $w$ joined to $a,b,c$ by three paths whose interiors are mutually disjoint and avoid $C$. The three arcs of $C$ between $a,b,c$, together with these three paths, form a subdivision of the complete graph on branch vertices $w,a,b,c$. Hence

A 3-connected graph need not contain a $T{K}_5$. The complete graph $K_4$ itself is 3-connected and has only four vertices, so it cannot contain a graph subdivision of $K_5$. Equally, any planar 3-connected graph is a counterexample because a subdivision of the complete graph $K_5$ would be nonplanar.