Let $I\subset \mathbb{R}$ be an interval and $f\in C^k(I, \mathbb{R})$, i.e a $k$-times continuous real valued function on $I$. Let $\mathbb{P}^N(I, \mathbb{R})$ be the set of polynomials of degree $N$ in $I$. Now by the Weierstrass Theorem we know that
$$ \forall \epsilon>0 \exists N, p\in\mathbb{P}^N(I, \mathbb{R}) \text{ s.t. } \sup_{t\in I} |p(t)-f(t)| < \epsilon $$
My question is if there exists another $p_k$ such that
$$ \sup_{t\in I} |p_k(t)-f(t)| +\sup_{t\in I} |p'_k(t)-f'(t)| + \ldots + \sup_{t\in I} |p^{(k)}(t)-f^{(k)}(t)| < \epsilon, $$ where the prime symbol indicates the first derivative and the $(k)$ super-index indicates the $k$-derivative.
Let $E\subset \mathbb R$ be a compact set and let $f\in C^k(E,\mathbb R)$. Because of our hypothesis on $f$, we know that $f^{(k)}$ is continuous. Hence , $\forall\epsilon>0$ we can find a polynomial $q$ such that $\Vert q-f^{(k)}\Vert<\epsilon$, where $$\Vert g\Vert:=\sup_{x\in E}|g(x)|$$ Is the supremum norm. Now let $a=\inf E$ and introduce the integral operator $$(\mathrm I\phi)(x)=\int_a^x \phi(s)\mathrm ds$$
Let $p=\mathrm I^k q$ so we can equivalently say $\Vert p^{(k)}-f^{(k)}\Vert < \epsilon$. Now let $\phi =p-f$ so we can equivalently say $$\Vert\phi^{(k)}\Vert < \epsilon$$ Of course, this means that $$\left|(\mathrm I\phi^{(k)})(x)\right|=\left|\int_a^x \phi^{(k)}(s)\mathrm ds\right|\leq \epsilon (x-a)$$ Letting $L$ be the measure (length) of $E$, we know then that $$|(\mathrm I\phi^{(k)})(x)|\leq \epsilon L \\ \forall x\in E$$ Hence $\Vert \mathrm I \phi^{(k)}\Vert=\Vert \phi^{(k-1)}\Vert \leq \epsilon L$. You can of course carry on this reasoning to say $\Vert\phi^{(k-m)}\Vert \leq \epsilon L^m$. Therefore $$\Vert \phi \Vert+ \cdots +\Vert \phi^{(k)}\Vert\leq \epsilon (L^k+\cdots +L^0)$$ Of course $L^k+\cdots +L^0=\frac{L^{k+1}-1}{L-1}$ and so if you choose our $q$ such that $\Vert q-f^{(k)}\Vert <\epsilon\frac{L-1}{L^{k+1}-1}$ then you get your desired bound.