Proving a set is closed in a metric space 
Given the subset 
  $$
F=\left\{f \in C[0,1]: \int_0^1 f(t)dt=1\right\}
$$
  show that $F$ is closed in $C[0,1]$ with the supremum metric. 

Definitions we use:
Limit point: $x$ is a limit point of $F$ if each open ball centered at $x$ contains at least one point of $F$ different from $x$, i.e. $S(x,r)-\{x\}$ intersects $F$.
Closed: a subset $F$ of a metric space is closed if it contains each of its limit points.
My brain is completely frozen on this question, and my apologies for the poor form of my entries this is my first time posting. Any help on this one would be greatly appreciated.
 A: I guess this here is more a job for something else, there is a definition of continuousity that states, a function is continuous when preimages of closed sets are closed. This fact is more likely to help us how you will see.
Integration is a bounded linear operator and hence continuous. Your set is the Preimage of $\{1\}$ of the integral function. What do you know about the preimage of closed sets of continuous functions?
To make it more explicit define 
\[h(f)= \int_0^1 f(x) \; \mathrm{d}x\]
As \[\|h(f)\| \leq \sup_{t\in [0,1]} |f(x)| \]
and $h(\alpha f + \beta g)=\alpha h(f) +\beta h(g)$ we see $h$ is a linear and bounded operator, hence $h$ is continuous. As preimages of closed sets are closed and 
\[ F=h^{-1} ( \{ 1 \} )\]
Another proof of the continuousity of the Integral. 
As we know the fundamental theorem of calculus   \[ \int_x^{x+h} f(s) \; \mathrm{d}s =h \cdot f(\xi) \] 
with $\xi \in (x,x+h)$ we gonna prove the Lipschitzcontinuouity of the Integral:
\begin{align*}
\left| \int_0^1 f(s)\; \mathrm{d}s - \int_0^1 g(s) \; \mathrm{d}s \right|&= 
\left| \int_0^1 f(s)-g(s) \; \mathrm{d}s \right|\\
&= \left| f(\xi)-g(\xi) \right| \leq \sup_{\xi \in [0,1]} |f(\xi)-g(\xi)|\\
&=1 \cdot \| f-g\| 
\end{align*}
So we have shown that 
$$ \|h(f)-h(g)\|_{2} \leq L \cdot \|f-g\|_{\text{sup}} $$
