This is example 5.5.4 from Elementary Classical Analysis, Marsden and Hoffman.
Let $$B = \{ f :[0,1] \rightarrow \mathbb{R} \; ; \; f \text{ is continuous, and } f(x) > 0 \; \forall \; x \in [0,1] \;\}. $$
The question is what is the closure of $B$.
( $\mathcal{C}$ is the set of continuous functions.)
Intuitively, it makes sense that the closure is $$D = \{ f \in \mathcal{C}; f(x) \geq 0 \; \forall \; x \in [0,1] \;\}. $$
But, the next sentence shows that this is closed since if a sequence of functions converges, the limit is in $D$.
Aren't we done? Why do we need to show that "for $f \in D$, there is an $f_n \in B$ such that $f_n \rightarrow f$? (That proves that all functions/elements of $B$ are limit points, but the definition of closure that I know is that the set contains all limit points, not that all points are limit points)
A set containing its limit points is closed. To be the closure of a given set, it has to be the smallest closed set containing it. That is why you need to verify that any point in $D$ can actually be obtained as a limit.