Proof:(By Contradiction)
On the contrary, a quadratic equation can have more than two roots. Suppose $\alpha$, $\beta$ and $\gamma$ are three distinct roots of a quadratic equation. Then these three roots must satisfy the quadratic equation. So, we can write it as the following:
$ a\alpha^{2}+b\alpha+c=0 ...(1)$
$a\beta^{2}+b\beta+c=0 ...(2)$
$a\gamma^{2}+b\gamma+c=0 ...(3)$
$(1)-(2):$ we have
$a(\alpha^{2}-\beta^{2})+b(\alpha-\beta)=0$
$(\alpha-\beta)\{a(\alpha+\beta)+b\}=0$
Since $\alpha-\beta≠0$. Therefore, $a(\alpha+\beta)+b=0 ...(4)$
$(2)-(3):$ we have
$a(\beta^{2}-\gamma^{2})+b(\beta-\gamma)=0$
$(\beta-\gamma)\{a(\beta+\gamma)+b\}=0$
Since $\beta-\gamma≠0$. Therefore, $a(\beta+\gamma)+b=0 ...(5)$
$(4)-(5):$ we have $a(\alpha-\gamma)=0$ Since $a≠0$. Therefore, $\alpha-\gamma=0 \implies \alpha=\gamma$. This is a contradiction to that fact that $\alpha$, $\beta$ and $\gamma$ are distinct. This establishes the proof.
Is this correct approach to prove it?
The comments immediately following the question discuss the OP's analysis, which I (also) consider valid. I favor the following alternative approach:
Suppose $f: \Bbb{R} \to \Bbb{R}$ has form
$a_n x^n + a_{n-1}x^{n-1} + \cdots a_1 x^1 + a_0 ~: ~a_n \neq 0.$
That is, $f(x)$ is an $n$-th degree polynomial.
Suppose further that $f(r) = 0$.
Then, there exists a polynomial $g(x)$ of degree $(n-1)$ such that $f(x) = (x-r)g(x).$
Proof
For $s \in \Bbb{Z^+}, ~~x^s - r^s = $
$(x - r) \times (x^{s-1} + x^{s-2}r + x^{s-3}r^2 + \cdots + xr^{s-2} + r^{s-1})$.
Let $h(s,r)$ denote the polynomial of degree $(s-1)$ that represents $\dfrac{x^s - r^s}{x - r}.$
Since $f(r) = 0,$ you have that
$\displaystyle f(x) = f(x) - f(r) = \sum_{k=1}^n \left[(x - r) \times a_k \times h(k,r)\right].$
Since each term has the common factor of $(x - r)$, this implies that $f(x) = f(x) - f(r) = $
$\displaystyle (x - r) \left\{\sum_{k=1}^n \left[a_k \times h(k,r)\right]\right\}.$
In the above summation, only the term that represents $(k = n)$ will feature a polynomial of degree $(n-1)$. Therefore, the overall summation must represent a polynomial $g(x)$ of degree $(n-1)$.
Therefore $f(x) = (x-r)g(x).$