I think the following proof works, ignore if you didn't want to see attempts at a proof but only regarding the question about $C_5$ specifically:
You can prove this by induction. Any connected simple graph of 2 vertices is a biclique. Suppose that for k <= n, any connected simple graph with k vertices not having $P_4$ or $C_3$ as an induced subgraph has this property. Now consider any graph of size n+1 that does not have either $P_4$ or $C_3$ as an induced subgraph. Then for any vertex v, the subgraph induced on all other vertices is a biclique (since it does not have $P_4$ or $C_3$ as an induced subgraph). Since the graph (call it G) is connected, there is a vertex w which is adjacent to v (in the induction case n>2). Call the partition of vertices to which w belongs in the n-sized induced subgraph (which is a biclique) $p_1$ and the other one $p_2$. Consider any vertex x in $p_2$. Since there is a vertex wx (by induction hypothesis), vx does not exist (else $C_3$ would become an induced subgraph). For any vertex other than v in $p_1$ (say y), xy also exists. If vy does not exist, then the path vwxy would induce $P_4$ in the subgraph of G consisting of v, w, x, y. So vy must also exist.
EDIT: As joriki pointed out, the subgraph induced on the rest of the vertices need not be connected. In that case, all we know for now is that each connected component of the subgraph induced on the rest of the vertices is a biclique, with the added condition that for each of these components there exists a vertex which has an edge to v.
Let $G_1$ and $G_2$ be two such components, with vertex w and x repectively that is connected to v. If removal of v splits G into isolated points, then G is already a biclique (one partition being v itself and other one being rest of vertices). Else, at least one out of $G_1$ and $G_2$ has at least one more vertex in addition to the one that is connected to v. Without loss of generality, we can assume this is $G_1$, and let the vertex be y. Now we know that ywvx is a connected path. yx and wx cannot exist as x is on a different connected component from y and w. vy does not exist as then ywv will be a triangle. So the subgraph og G induced on y,w,v,x will then be $P_4$. So we can safely rule out this situation too.