When is the pointwise convergence on $A$ equivalent to the uniform one? Let $A$ be a finite set of $[0,1]$ and $F$ a sub-vector-space of $C([0,1],\mathbb R^n)$.
When is the pointwise convergence on $A$ equivalent to the uniform one.
And with $A$ be countable?
I suppose the finite dimension of $F$ is sufficient but is it necessary? Not sure for the second case.
Is there a more general condition to have this equivalence (with any vector-space)?
 A: In this answer I assume  the second interpretation from my comment. In the thread linked there I answered the question for the case of finite $A$ and $n=1$. I expect the case of $n>1$ can be dealt very similarly.
I can add to that my answer the following.
There exists a sequence $f_n$ of functions of $C([0,1],\Bbb R)$ which converges on $[0,1]$ to the zero function pointwise, but not uniformly. Namely, for each $n$ let $f_n(0)=f_n(1)=f_n(1/n)=0$, $f_n(1/(2n))=1$ and between the points $0$, $1/(2n)$, $1/n$ and $1$ the function $f_n$ is linear. 
For each $A$ not dense in $X$ we can pick a non-zero function $f_A\in C([0,1],\Bbb R)$ such that $f_A|A$ is zero. Let $F$ be a one-dimensional linear space $F\subset C([0,1],\Bbb R)$ spanned by $f_A$. Then $f|A$ is zero function for any $f\in F$, so pointwise convergence on $A$ to zero of the sequence $\{f_n\}$ of functions of $F$ does not imply its (uniform) convergence on $[0,1]$. 
For any linear space $F\subset C([0,1],\Bbb R)$, whose dimension $\dim F$ is finite there exists a subset $A$ of $[0,1]$ such that $|A|=\dim F $ and restriction $f|A$ of $f$ on $A$ is a non-zero function for each non-zero function $f\in F$. So, in this case, by proposition from my answer,  for each sequence $\{f_n\}$ of functions of $F$ and each function $f\in F$, $\{f_n\}$ converges to $f$ uniformly on $[0,1]$ iff $\{f_n|A\}$ converges to $f|A$ pointwise on $A$.
The existence of such a set $A$ we can prove by induction with respect to $\dim F$. For $\dim F=1$ pick $A=\{a\}$, where $a\in [0,1]$ is an arbitrary point such that $f(a)\ne 0$ for any non-zero function $f\in F$. Assume that we alredy have proved the induction hypothesis for $\dim F\le d$. Let $\dim F=d+1$ and $b_1,\dots, b_{d+1}$ be the basis of the space $F$. Let $F’$ be the linear subspace of $F$ spanned by functions $b_1,\dots, b_d$. By the induction hypothesis there exists a subset $A’$ of $[0,1]$ of size $d$ such that $f|A’$ is non-zero function for each non-zero function $f\in F’$. Put $G=\{f\in F:f|A’$ is the zero function$\}$. Since $|A’|=d$, the map $F’\to\Bbb R^{A’}$, $f\mapsto f|A$ is surjective. Using this we can easily show that $F=F’+G$. Now, let $f, g\in G$ be any functions. There exist not both zero $\lambda$ and $\mu$ such that $\lambda f+\mu g\in F$. Since  $(\lambda f+\mu g)|A=0_A$, $\lambda f+\mu g=0$. Thus $\dim G=1$. Pick an arbitrary point $a\in [0,1]$ such that $f(a)\ne 0$ any non-zero function $f\in G$ and put $A=A’\cup\{a\}$. Using that $F=F’+G$, we can easily show that $A$ is a required set. 
But I don’t know whether there exists an infinitely dimensional $F$ such that pointwise convergence on $[0,1]$ to zero of the sequence $\{f_n\}$ of functions of $F$ implies its uniform convergence.
