http://www.artofproblemsolving.com/Forum/viewtopic.php?p=114549&sid=dc4938f299313a92d2edcabcb904ed75#p114549

This problem is called as Happy End Problem which Erdos named. We consider the full convex of $5$ points which we call $CONV$. Case1:$CONV$ is pentagon ABCDE ABCD satisfies the condition. Case2:$CONV$ is quadrilateral ABCD ABCD satisfies the condition. Case3:$CONV$ is triangle ABC $2$ points D,E is inside of ABC.WLOG straight line DE meets AB and CD.Then BCDE satisfies the condition. From case1,2,3,the proof is completed.$\blacksquare$

Compare with 1966 ILL #4. Let the $5$ points be $A$, $B$, $C$, $D$, and $E$. We first consider $A$, $B$, $C$, and $D$. If each of these is in the exterior of the triangle determined by the other $3$ points, we are done by $ABCD$. If one of these is in the interior, however, WLOG let it be $A$. We now proceed with casework based on the position of $E$. If $E$ is in the exterior of triangle $BCD$, the lines $AB$, $AC$, and $AD$ split the plane into $3$ regions. If $E$ is in the region determined by $AB$ and $AC$, then $ABEC$ is convex, and analogous reasoning applies to the other $2$ regions. If $E$ is in the interior of triangle $BCD$, extend $BA$ bast $A$ to intersect $CD$ at $E$, and define $F$ and $G$ similarly. Then, if $E$ is in the interior of triangle $ACG$ or $ADF$, $AECD$ or $AEDC$ is convex respectively. Analogous reasoning applies to the other $2$ pairs of triangles. $\blacksquare$