If $A$ has pairwise distinct diagonal elements one can prove the theorem by induction as follows:
- By block partitioning one can put $A\in T_{n}({\mathbb F})$ to the form $$ A = \begin{pmatrix} \alpha & a \left(\alpha I - A'\right)\\ 0 & A'\end{pmatrix}$$ with $\alpha \in{\mathbb F}$, $a\in {\mathbb F}^{n-1}$ a row vector and $A'\in T_{n-1}({\mathbb F})$. Here one uses that $\alpha I - A'$ is non singular. For $p \in {\mathbb F}[t]$ one has $$ p(A) = \begin{pmatrix} p(\alpha) & a \left(p(\alpha) I - p(A')\right)\\ 0 & p(A')\end{pmatrix}.$$
- For $X \in T_{n}({\mathbb F})$, one has $[A,X]=0$ iff $X$ can be put to the form $$ X = \begin{pmatrix} \xi & a \left(\xi I - X'\right)\\ 0 & X'\end{pmatrix}$$ with $[A',X']=0$.
- Now, if $B\in T_{n}({\mathbb F})$ satisfies $[B,X]=0$ it must be (by Step 2) of the form $$ B = \begin{pmatrix} \beta & a \left(\beta I - B'\right)\\ 0 & B'\end{pmatrix},$$ where $[B',X']=0$. By induction hypothesis, there is a polynomial $q\in {\mathbb F}[t]$ such that $B'=q(A')$.
- Now, if one denotes the pairwise distinct diagonal elements (i.e., eigenvalues) of the upper triangular matrix $A$ by $\alpha, \lambda_1,\ldots,\lambda_{n-1}$ the polynomial $p \in {\mathbb F}[t]$ solving the interpolation problem $p(\alpha)=\beta$ and $p(\lambda_j)=q(\lambda_j)$ ($j=1,\ldots,n-1$) yields $p(A')=q(A') = B'$ and hence, by Step 1, $p(A)=B$.
In particular, the thus constructed polynomial $p$ has degree $n-1$.
For ${\mathbb F} ={\mathbb R}$ or ${\mathbb C}$ one could use continuity arguments to show the general case.