The question is
Let $f:[0,1]\to\mathbb{R}$ be a continuous function. Let $\varepsilon>0$, show that there is a polynomial $p:[0,1]\to\mathbb{R}$ with $\vert f(x)-p(x)\vert <\varepsilon$ for all $x\in[0,1]$ and with $p(0)=f(0),p\left(\frac{1}{2}\right)=f\left(\frac{1}{2}\right), p(1)=f(1)$
What I have in my mind is that using the Weierstrass approximation theorem, let $q:[0,1]\to \mathbb{R}$ such that $$ \forall \varepsilon>0, \vert f(x)-q(x)\vert <\frac{\varepsilon}{4} $$ Then $\vert f(x)-p(x) \vert \leq \vert f(x)-p(x) \vert+\vert f(0)-p(0) \vert + \vert f\left(\frac{1}{2}\right)-p\left(\frac{1}{2}\right)\vert + \vert f(1)-p(1) \vert <\frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{\varepsilon}{4}=\varepsilon$. However, the equality conditions are not satisfied.
For only one point in the equality conditions, it is easy to see that the above argument works, changing the denominator of $\varepsilon$ to 2 is enough. But I got stuck in this problem. Do I have to consider some fancy linear combination of the three points? Or is there any other ways to show it? Many thanks!
Hint: Let $q$ be a polynomial such that $|f(x)-q(x)| <\epsilon $ for all $x$. Let $p(x)=q(x)+a+bx+cx^{2}$. We can choose $a,b,c$ so that $f(0)=p(0),f(\frac 1 2)=p(\frac 1 2)$ and $f(1)=p(1)$. Now use the fact that $|f(x)-q(x)| <\epsilon $ for $x=0,\frac 1 2$ and $x=1$ to check that $|a|<\epsilon, |a+\frac b 2 +\frac c 4| <\epsilon$ and $|a+b+c| <\epsilon$. Conclude that $|a|, |b|, |c|$ are all 'small' to finish the proof.
[You should check that $|a|<\epsilon,|b| <10\epsilon $ and $|c| <12 \epsilon$. Finally you get $|f(x)-p(x)| <36 \epsilon$ for all $x$].