Asked here before. But I had the idea to try and prove this using Rolle's Theorem and Induction. This is certainly not the easiest way to prove this, as seen in the link above. However, I am still new to analysis and I am trying to determine if what I have done actually works, I am a little hung up on how to prove non-distinct zeroes, but other than that I think what I have here is good. Here it is:
(Rolle's Theorem) If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and $f(a)=f(b)$ then there is a point $c\in(a,b)$ so that $f'(c)=0$.
(Exercise 4.4) Every polynomial is a continuous function.
(Exercise 5.1) If $f(x)=x^n$, where $n\in\mathbb{N}$ then $f'(x)=nx^{n-1}$.
Proof. Let $P_n=a_0+a_1x+\ldots+a_nx^n$ be any $n$ degree polynomial where $n\in\mathbb{N}$ and $a_i\in\mathbb{R}$. By Exercises 4.4 and 5.1 and the arithmetic rules for limits, we know every polynomial is both continuous and differentiable on $\mathbb{R}$. By Exercise 5.1 and the arithmetic rules for limits, we know that the derivative of a polynomial is a polynomial. Therefore, for $n\geq k$, where $k\in\mathbb{N}$ the $k$th derivative of $P_n$, that is $P_n^{(k)}$, is continuous and differentiable on $\mathbb{R}$.
Suppose $P_n$ has $n+1$ distinct real zeroes $x_1<\ldots<x_{n+1}$. Then $P_n(x_1)=\ldots=P_n(x_{n+1})=0$. Since $P_n$ is continuous and differentiable for all $x\in\mathbb{R}$, $P_n$ is also continuous on $[x_i,x_{i+1}]$ and differentiable on $(x_i,x_{i+1})$ for each $i\in\{1,\ldots,n\}$. By Rolle's Theorem $\exists \ c_{1i}\in(x_i,x_{i+1})$ so that $P_n'(c_{11})=\ldots=P_n'(c_{1n})=0$. Then $P_n'$ has $n$ (that is $n+1-1$) distinct real zeroes where $c_{11}<\ldots<c_{1n}$.
Suppose $P_n^{(k)}$ has $n+1-k$ distinct real zeroes $c_{k1}<\ldots<c_{k(n+1-k)}$. Then $P_n^{(k)}(c_{k1})=\ldots=P_n^{(k)}(c_{k(n+1-k)})=0$. Since $P_n^{(k)}$ is continuous and differentiable on $\mathbb{R}$, we know it is continuous on $[c_{ki},c_{k(i+1)}]$ and differentiable on $(c_{ki},c_{k(i+1)})$ for each $i\in\{1,\ldots,n-k\}$. By Rolle's Theorem $\exists \ c_{(k+1)i}\in(c_{ki},c_{k(i+1)})$ such that $P_n^{(k+1)}(c_{(k+1)1})=\ldots=P_n^{(k)}(c_{(k+1)(n-k)})=0$. Then $P_n^{(k+1)}$ has $n-k$ (that is $(n+1)-(k+1)$) real zeroes. Thus by induction, if $P_n$ has $t$ distinct real zeroes, $P_n^{(k)}$ has $t-k$ distinct real zeroes.
Then we have that $P_n^{(n-1)}$ has $(n+1)-(n-1)=2$ distinct real zeroes $c_{(n-1)1}<c_{(n-1)2}$. So $P_n^{(n-1)}(c_{(n-1)1})=P_n^{(n-1)}(c_{(n-1)2})=0$. By Exercise 5.1 and arithmetic rules for limits we find that $P_n^{(n-1)}(x)=(n-1)!a_{n-1}+n!a_nx$. So it must be that $c_{(n-1)1}=c_{(n-1)2}$, a contradiction. Therefore, it must be that $P_n$ has at most $n$ real zeroes. $\square$
I now have a much simpler way to do this still using Rolle’s Theorem:
Proof. Let $P_n$ be a non-zero degree $n$ polynomial. By Exercises 4.4 and 5.1 and the arithmetic rules for limits, we know every polynomial is both continuous and differentiable on $\mathbb{R}$. By Exercise 5.1 and the arithmetic rules for limits, $P_{n+1}'=P_n$. Since $P_0$ is non-zero, for some $a\in \mathbb{R}\backslash\{0\}$, $P_0(x)=a$ for all $x\in \mathbb{R}$. So there is no $x\in \mathbb{R} $ having the property $P_0(x)=0$, so $P_0$ has at most 0 zeroes. Suppose $P_k$ has at most $k$ real zeroes, and $P_{k+1}$ has more than $k+1$ real zeroes, say $k+2$ real zeroes $x_1<\ldots<x_{k+2}$. Since $P_{k+1}$ is continuous on $[x_i,x_{i+1}]$ and differentiable on $(x_i,x_{i+1})$ for each $i\in J=\{1,\ldots,k+1\}$, and $P_{k+1}(x_1)=\ldots=P_{k+1}(x_{k+2})=0$, by Rolle's Theorem $\exists \ c_i\in(x_i,x_{i+1})$ such that $P_{k+1}'(c_i)=0$ for each $i\in J$. So $P_{k+1}'$ has $k+1$ real zeroes, however $P_{k+1}'=P_k$ which has at most $k$ real zeroes, a contradiction. Therefore by induction $P_n$ can have at most $n$ real zeroes. $\square$