How to prove that if the determinant of the matrix is zero then at least one eigenvalue must be zero?

Let $p(x)=\det(A-xI)$ the char. polynomial of $A$. Then $p(0)=\det(A)=0$, hence $0$ is a root of $p$ and therefore an eigenvalue of $A$.

Here an elementary way:

$\det(A) = 0 \Rightarrow$ the columns of $A =(c_1 \ldots c_n)$ are linearly dependent $\Rightarrow$ there is a non-zero vector $v = (v_1 \ldots v_n)^T$ such that $v_1c_1 + \cdots v_n c_n = \vec{0} \Rightarrow Av = \vec{0} = 0\cdot v \Rightarrow 0$ is an eigenvalue of $A$.

I do not know what you know about the determinant and how you think of it, but the determinant of a square matrix $A$ is zero iff the matrix is not invertible, and that is equivalent to the kernel being non-trivial, which means that $Ax=0$ for some $x\ne0$.