A graph is planar if there exist at least one drawing of this graph (called an embedding) on the plane with no crossing edges.
Therefore as explained by Bercy, you just need to show one drawing of this graph with no crossing edges. To do so take you vertex labelled 0, and pull it out of the other vertices, outside the star-like graph around the vertex 4. Remember that edges are not required to be straight line (even if there is some result proving that this is always possible). So you can use bent lines to join some vertices. You should be able to build such a planar drawing.
Edit I just noticed your other question here. There you were asked to prove that a graph is not planar. That's way you needed to prove that it includes $K_5$ or $K_{3,3}$ as a subgraph. Because otherwise you would need to exhibit all possible drawings. However, here to prove that a graph is planar, it is enough to exhibit one planar drawing.