Suppose $\{f_n\}$ is a sequence of functions on $\mathbb R$ and that for any $a,b\in\mathbb R$, the sequence of restricted functions $f_n |_{[a,b]}\rightarrow f_{[a,b]}$ uniformly. (My notation here is not great).
Is it possible to use the $f_{[a,b]}$ and piece them together to find a function $f$ such that $f_n \rightarrow f$ uniformly on $\mathbb R$?
I think it's true that $f_{[a,b]}$ and $f_{[c,d]}$ have to agree on the intersection of $[a,b]$ and $[c,d]$. Not sure whether this is enough though. It's reminding me a lot of presheaves.
Is this possible? How does one (dis)prove it?
No it is not possible. The uniform convergence on compact intervals is a very common phenomenon, take for example the sequence $$ f_n(x)=\frac{x}{n},\qquad x\in\mathbb R.$$ Since $$ |f_n(x)|\le \frac{M}n,\qquad |x|\le M, $$ the sequence converges uniformly to $0$ on $[-M, M]$, but it cannot converge uniformly on $\mathbb R$, because $$ \sup_{x\in \mathbb R} |f_n(x)|=\infty,\qquad \forall n\in\mathbb N.$$
Another common occurrence of the phenomenon is with power series. The sequence $$ f_n(x)=\sum_{k=0}^n \frac{x^k}{k!}, $$ for example, does converge uniformly on every bounded interval, but it does not converge uniformly on $\mathbb R$.