Step 1: We have for all $p(t)=a_0\cdot t^0+a_1\cdot t^1+\cdots+a_n\cdot t^n$ that $ \int_{0}^{1} p(xt)\,dx= \int_{0}^{1} a_0\cdot (xt)^0+a_1\cdot (xt)^1+\cdots+a_n\cdot (xt)^n \,dt =0 $ for all $x\in\mathbb{R}$ if, only if, $a_0=0,a_1=0,\dots,a_n=0$. In fact, $ \int_{0}^{1} p(tx) dt = a_0 x +\frac{1}{2}a_1x^2+\frac{1}{3}a_2x^3+\dots+\frac{1}{n+1}a_nx^{n+1}=0 $ for all $x\in\mathbb{R}$. Now use induction on $n\in\mathbb{N}$ and conclud that $a_1=0,a_2=0,\dots a_n=0$.
Step 2: use the classical Weierstrass aproximation theorem
Theorem ( Weierstrass aproximation) For all continuous fuction $f$ defined in a closed interval $[a,b]$ there is a sequence $\{ p_n(\cdot)\}_{n\in\mathbb{N}}$ of polinoms shout that $\lim_{n\to\infty}p_{n}(x)=f(x)$. And the convergence of $\lim_{n\to\infty}p_{n}(x)$ is uniform.
and a integral lema
Lema. Let's $f$ and $f_n$, wthi $n\in\mathbb{N}$, functions in a clused interval $[a,b]$. If the convergence of $\lim_{n\to\infty}f_{n}(x)=f(x)$ is uniform them $ \lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dx=\int_{a}^{b}\lim_{n\to\infty} = f_{n}(x)\,dx=\int_{a}^{b} f(x)\, dx $