Let $C^{1}[0,1]$ denote the set of once-continuously differentiable functions $f:[0,1]\rightarrow\mathbb R$. For every $f\in C^1[0,1]$, find a polynomial $p$ such that $\sup_{x\in[0,1] }|f(x)-p(x)|<\epsilon$ and $\sup_{x\in[0,1] }|f'(x)-p'(x)|<\epsilon$.
So, I can find different polynomials for both, by the Stone-Weirstrass theorem. What I am missing is how to connect them: how can I make sure that the polynomial $p$ I find, or a polynomial $p'$ that I find for the second condition, satisfy the other condition? It really feels like there should be a simple trick to create a third function from some other information.
Since we know that $f\in C^1[0,1]$, if follows that $f'\in C[0,1]$ and we can find a polynomial $\tilde p$ such that $\sup_{x\in [0,1]} |f'(x)-\tilde p(x)| \le \varepsilon$.
Let $p$ be the antiderivative of $\tilde p$ such that $p(0)=f(0)$, i.e. $$ p(x):=f(0) + \int_0^x\tilde p(t)\, dt. $$ It is obvious that $p$ is a polynomial and that $\sup_{x\in [0,1]} |f'(x)-p'(x)| \le \varepsilon$.
Now, we have $$\begin{align} f(x)-p(x) &= \left(f(0) + \int_0^x f'(t)\, dt \right)-\left(f(0) + \int_0^x p'(t)\, dt \right) \\ &=\int_0^x f'(t)-p'(t)\, dt, \end{align}$$ hence we can take absolute value and get $$\begin{align} |f(x)-p(x)| &\le\int_0^x |f'(t)-p'(t)|\, dt \\ &\le \varepsilon \int_0^x 1 \,dt \\ &\le \varepsilon \end{align}$$ for all $x\in [0,1]$. This proves what you want.