Checking the uniform denseness of a set in $C([0, 1], \mathbb{R}^2)$ 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?
 A: $\newcommand\la\lambda\newcommand\R{\mathbb R}$It is true that, if $S_2$ is dense in $C([0,1],\R^2)$ with respect to the sup-norm $\|\cdot\|$, then $S_1$ is uniformly dense in $C([0,1],\R^2)$ as well.
However, the reason for this may disappoint you: it is that $S_2$ is actually never dense in $C([0,1],\R^2)$ with respect to $\|\cdot\|$ -- for any continuous functions $\la$ and $b_{1j},b_{2j}$ with $j\in[m]:=\{1,\dots,m\}$.
Indeed, there are two cases to consider:
Case 1: $\la(t)=c$ for some real $c$ and all $t$. Then $S_2$ is finite dimensional and therefore cannot be dense in $C([0,1],\R^2)$.
*Case 2: there are $t_1$ and $t_2$ such that $0\le t_1<t_2\le1$ and $\la(t_1)=:u\ne v:=\la(t_2)$. Let $x_2(t):=1$ for $t\in[0,1]$. Take any real
$$w>\max_{j\in[m]}\Big(2\|b_{1j}\|+\frac4{|v-u|}\,\|b_{2j}\|\Big) \tag{0}$$
and let  $x_1(t):=w$ for $t\in[0,1]$.
Suppose the contrary to our claim: that $S_2$ is dense in $C([0,1],\R^2)$. Then for each natural $n$ there exist a polynomial $P_n$ and some $j_n\in[m]$ such that
$$P_n(\la)b_{1j_n}+P'_n(\la)b_{2j_n}\to w,$$
$$P_n(\la)b_{2j_n}\to1,$$
where the convergence is uniform. Passing to a subsequence, without loss of generality (wlog) we may assume that $j_n$ does not depend on $n$, and then wlog $j_n=1$ for all $n$. For brevity, let then $b_1:=b_{11}$ and $b_2:=b_{21}$. So,
$$P_n(\la)b_1+P'_n(\la)b_2\to w,\tag{1}$$
$$P_n(\la)b_2\to1.\tag{2}$$
Since the convergence is uniform, it follows from (2) that the function $b_2$ does not take the zero value on $[0,1]$. Therefore and because $b_2$ is continuous, wlog $|b_2|\ge1$, whence, again by (2),
$$\|P_n(\la)\|\le2 \tag{4}$$
eventually (for all large enough $n$).
So, by the mean value theorem and the intermediate value theorem, eventually
$$4\ge|P_n(\la(t_2))-P_n(\la(t_1))| \\ 
=|P'_n(\la(s_n))|\,|\la(t_2)-\la(t_1)| \\ 
=|P'_n(\la(s_n))|\,|v-u| \tag{5}$$
for some $s_n\in(t_1,t_2)$, whence, in view of (4), (5), and (0),
$$P_n(\la(s_n))b_1(s_n)+P'_n(\la(s_n))b_2(s_n)
\le2\|b_1\|+\frac4{|v-u|}\,\|b_2\|< w,$$
which contradicts the uniform convergence in (1). $\Box$
