If $B$ is invertible then there exists a scalar $c$ such that $A+cB$ is not invertible

Let's take the full journey. $B$ is inversible so $B^{-1}$ exists and $\det{B}\neq 0$. So we can rewrite

$$\det(A+cB)=\det(AB^{-1}+cI)\det{B}$$

Now the coefficient of $c^n$ in the $\det(AB^{-1}+cI)$ is $1$ so the polynomial has degree $n\gt 1$ and so it has a root in $\Bbb{C}$.

Now let's find a counterexample. Consider

$$ A=\begin{bmatrix} 1 & 0\\1 & 1\end{bmatrix}$$

$$ B=\begin{bmatrix} 0 & 1\\0 & 1\end{bmatrix}$$

One has

$$A+cB =\begin{bmatrix}1 & c\\1 & 1+c\end{bmatrix}$$

Hint: Consider the polynomial $\varphi(z) = 1$ in $\mathbb C$. Does it have a root?