Intuition on an unexpected finite dimensional space

Perhaps surprising but it is in fact possible to get an explicit bound for the dimension of $F$ (math magic at work). This is a bit like a constructive proof of Riesz's theorem.

We construct recursively subspaces of $F$: Let $f_1\in F$ and $x_1\in [0,1]$ be such that $f_1(x_1)=1=\|f_1\|_\infty$ and define $$ F_1 = \{ f\in F : f(x_1)=0 \}$$ If $F_1$ is non-trivial find $f_2\in F_1$ and $x_2\in [0,1]$ (necessarily different from $x_1$) such that $f_2(x_2)=1=\|f_2\|_\infty$. Then set $$ F_2=\{f\in F: f(x_1)=f(x_2)=0\}$$ Proceed inductively as long as you can. If finally $F_{d+1}=\{0\}$ your space is of dimension $d$. Now each $f_k$ vanishes at $x_i$, $i<k$ and $f_k(x_k)=1=\|f_k\|_\infty$ and by the MVT we must have $$ C\geq \|f_k'\|_\infty \geq \max \{ \frac{1}{|x_k-x_i|} : 1\leq i<k\}$$ Therefore, $$ |x_k-x_i| \geq 1/C$$ for all $1\leq i<k \leq d$. This implies $d\leq C+1$.

I had this in the comments earlier, but when I went to add an example, I found it too frustrating to edit in the comment box.

Suppose that $C = 1$. So the condition of interest on a subspace $F$ is: $\|f'\| \leq \|f\|$ for all $f \in F$.

Since $(e^x)'=e^x$ and $(e^{-x})' = -e^{-x}$ it is easy to see that the lines $L_1 = \mathbb{R} \cdot e^x$ and $L_2 = \mathbb{R} \cdot e^{-x}$ both satisfy the condition. However, there cannot be a subspace $F$ satisfying the property which contains both $L_1$ and $L_2$. Such a space $F$ would need to contain $f = e^{x} - e^{-x}$, but one can check that $\|f'\| > \|f\|$, so this is impossible.

The conclusion from the above is that, for a given $C$, there is not just one example of a subspace $F$, but many. Moreover, there is not a largest one. That said, it might be interesting to ask if there is a finite bound on the dimensions of spaces satisfying the property for given $C$, and then search for examples realizing the maximal dimension.