So I have to prove that given a 3-dimensional compact polytope such that every two vertices are adjacent, then it is a tetrahedron.
Somehow I'm able to visualise from the fact that every two vertices are adjacent and sees that it fits to the diagram of tetrahedron. Like it is a polyhedron composed of four triangular faces, six straight edges, and four vertex corners. From this it is suffice that every other vertices is adjacent to each other. But I'm not quite sure how to prove it.
Could you please help me in proving the above statement.
Cheers.
As stated, this is false; the Császár polyhedron is a counterexample.
If your polyhedron is stipulated to be topologically equivalent to a sphere, then we may apply Euler's formula of $V-E+F=2$. Suppose that our polyhedron has $n$ vertices. Then it has$\frac{n(n-1)}2$ edges, and at most $\frac{2E}3$ faces, because each face touches at least three edges and every edge is shared by two faces.
So we have
$$n-\frac{n(n-1)}2+\frac{n(n-1)}3\ge V-E+F = 2$$
It is easy to check that only for $n=3,4$ is this inequality satisfied, and as the $n=3$ case yields a degenerate polyhedron (two triangles glued back-to-back), $n=4$ is the only possibility, and only works if each face is a triangle.