Condition for the variation of action is:
$$0=\delta S$$
$$=\int d^4 x [\frac{\partial \mathcal{L}}{\partial \phi}\delta\phi-\partial_\mu(\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)})\delta \phi+\partial_\mu (\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi)].$$
It is clear that the last summand is zero because $\delta \phi=0$ at the beginning (B) and ending (E):
$$\int_{x_B}^{x_E} d x (\frac{\partial \mathcal{L}}{\partial \dot\phi}\delta\phi)|_{t_B}^{t_E}+\int_{t_B}^{t_E} d t (\frac{\partial \mathcal{L}}{\partial \phi'}\delta\phi)|_{x_B}^{x_E}=0.$$
In every source I find they say only that $$\int d^4 x [(\frac{\partial \mathcal{L}}{\partial \phi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)})\delta \phi]$$ is zero if $$\frac{\partial \mathcal{L}}{\partial \phi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}=0.$$ That is obvious. But is it provable that $$\int d^4 x [(\frac{\partial \mathcal{L}}{\partial \phi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)})\delta \phi]\neq 0$$ for $$\frac{\partial \mathcal{L}}{\partial \phi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}\neq0~? $$
I tried to integrate this, for example with partial integration, but I don't get far with this and I don't have a clue how to proove this (if there is any proof).