Let $\lambda:[0, 1]\to \mathbb{R}$, and $b_{1j}, b_{2j}:[0, 1] \to \mathbb{R}$, $j = 1, \ldots, m$ be smooth functions. Consider the following two sets

$$\begin{align*} S_1 &= \left\{ \begin{bmatrix} P(\lambda)b_{1j} \\ P(\lambda)b_{2j} \end{bmatrix} \bigg\vert \:j = 1, \ldots, m, P(\lambda) \text{ is a polynomial of } \lambda\right\}\\ S_2 &= \left\{ \begin{bmatrix} P(\lambda)b_{1j} + P'(\lambda)b_{2j} \\ P(\lambda)b_{2j} \end{bmatrix} \bigg\vert \:j = 1, \ldots, m, P(\lambda) \text{ is a polynomial of } \lambda\right\}. \end{align*}$$

I'm stuck proving or disproving the following statement:

If $S_2$ is dense in $C([0, 1], \mathbb{R}^2)$ with respect to sup-norm, then $S_1$ is uniformly dense in $C([0, 1], \mathbb{R}^2)$ as well.

The major issue was that there is no guarantee that we can use $\{P_n\} \subset C^1([0, 1])$ to uniformly approximate $f\in C^1([0, 1])$ with $f'\in C([0, 1])$ be uniformly approximated by $\{P'_n\}\subset C([0, 1])$. I tend to believe the answer to the above statement is negative. But I cannot figure out a counter-example to it. Or is there any way to provide a positive answer to the statement?