Consider two matrices $A, B\in\mathcal{M}_n(\mathbb{C})$, such that $A, B$ has no common eigenvectors. Is it true that for some nonzero $t\in\mathbb{C}$, matrix $A+tB$ is similar to diagonal matrix $\text{diag}(a_1, a_2, \ldots, a_n)$, where $a_i\not = a_j, \forall i\not= j$?

It's false for every $n\ge 3$.

Consider $A_n,B_n$ $n\times n$ square matrices with no common eigenvalue (these exist for $n=0$ and $n\ge 2$). Then $$A=\begin{pmatrix} A_2 & 0 & 0\\ 0 & A_2 & 0\\ 0 & 0 & A_{n-4}\end{pmatrix},\qquad B=\begin{pmatrix} B_2 & 0 & 0\\ 0 & B_2 & 0\\ 0 & 0 & B_{n-4}\end{pmatrix}$$ work for $n=4$ and $n\ge 6$.

It remains to do $n=3,5$. Define $$N=\begin{pmatrix} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 &1& 0\end{pmatrix},\qquad N'=\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & -1\\ 0 &0& 0\end{pmatrix}$$ Then $N$ and $N'$ have no common eigenvector but the plane spanned by $N,N'$ consists of nilpotent matrices. Then the matrices $$A=\begin{pmatrix} N & 0\\ 0 & A_{n-3}\end{pmatrix},\qquad B=\begin{pmatrix} N' & 0\\ 0 & B_{n-3}\end{pmatrix}$$ work for all $n=3$ and $n\ge 5$.