Reduce probability space to the unit interval linear measure in the proof for Glivenko-Cantelli Theorem Glivenko-Cantelli Theorem states that:

Let $X_1,X_2,...$ be i.i.d. random variables and let
  \begin{align}
F_{n}(x) = n^{-1} \sum_{i=1}^{n} \mathbb{1}(X_{i}\leq x),
\end{align}
  Then 
  \begin{align*}
\sup_{x}|F_n(x)-F(x)| \to 0\  \text{a.s.} \text{ as}\ n \to \infty.
\end{align*}

I saw that in one proof the author first reduces the proof for the unit interval probability space with linear measure,  $([0,1],\mathcal{B}[0,1],\lambda)$. 

Let $\Lambda$ to be the CDF for $\lambda$ and suppose $Y_i$ are i.i.d. $\sim \lambda$. We define 
  \begin{align*}
\Lambda_n(F(x))=\frac{1}{n}\sum_{i=1}^n \mathbb{1}(Y_i\leq F(x)).
\end{align*}
  If $X_i= F^{-1}(Y_i)$, where $F^{-1}(y)= \inf \{x: F(x)\geq y\}$, then $X_i$ are i.i.d. with CDF $F$, and $X_i \leq x \Leftrightarrow Y_i \leq F(x)$. To prove the theorem, it suffices to show that 
  \begin{align*}
\sup_{y} |\Lambda_n(y)-\Lambda(y)|\to 0\ \text{a.s}
\end{align*}
  Fix $\epsilon >0$, choose $m$ s.t. $\frac{1}{m} \leq \frac{\epsilon}{2}$. Consider the set 
  \begin{align}
E=\{\frac{k}{m}: k=0,1,\ldots,m\}. 
\end{align}
  From SLLN, $\Lambda _n (y) \to \Lambda(y)$ for each $y\in E$. Since $E$ is finite, $\exists\ N$ s.t. $\forall\ n \geq N$, $|\Lambda_n (y)-\Lambda (y)| \leq \frac{\epsilon}{2} \ \forall\ y \in E$. For $x \in [0,1]$, we can always find $u,v$ s.t. $u\leq x <v$, $u,v \in E$, $v-u=\frac{1}{m}$.
  Since $\Lambda (y)=y$, we have
  \begin{align*}
\Lambda_n (x)\geq \Lambda_n (u) \geq u-\frac{\epsilon}{2} \geq x-\frac{1}{m} -\frac{\epsilon}{2} \geq x-\epsilon,\\
\Lambda_n (x)\leq \Lambda_n (v) \geq v+\frac{\epsilon}{2} \leq x+\frac{1}{m} +\frac{\epsilon}{2} \leq x+\epsilon,
\end{align*}
  so that
  \begin{align*}
|\Lambda_n(x)-\Lambda(x)|\leq \epsilon.
\end{align*}
  The theorem is now proved.

I don't see why it suffices to reduce the proof to the unit interval probability space case. Also, this proof technique seems quite common in probability theory. 
Assume the $X_i$ are defined on the same probability space $(\Omega, \mathcal{F}, \mathbb{P})$, why we can write $X_i= F^{-1}(Y_i)$ even $\mathbb{P}(Y_i\le F(x)) = \mathbb{P}(X_i\le x) = F(x)$. If we write $X_i$ in this way, now $X_i(\omega)$ is defined on $[0,1]$ instead of $\Omega$. Is this because every probability space can be viewed as a mapping from $([0,1],\mathcal{B}[0,1],\lambda)$? So $(\Omega, \mathcal{F}, \mathbb{P})$ can be viewed as a mapping from $([0,1],\mathcal{B}[0,1],\lambda)$, and eventually $X_i$ is defined on $[0,1]$. 
Even the above thinking is correct, then why with the same mapping $F^{-1}$, the proof for the unit interval probability space can be generalized to the general i.i.d. $X_i$. This mapping $F^{-1}$ is now taking  action on infinite many i.i.d. $Y_i$, how to write it as a form like
$\mathbb{P}(\sup_{x} |F_n(x)-F(x)| \ne 0) \le \mathbb{P}(\sup_{y} |\Lambda_n(y)-\Lambda(y)| \ne 0) = 0 $?
Can someone explain more here? I feel lost in thinking.
 A: The thing is: Your statement isn't really about the variables (as measurable maps), it's about their distributions (as measures on $\mathbb{R}$).
If you follow the reasoning from above, then any probability distribution $\mathbb{P}$ on $\mathbb{R}$ with CDF $F$ can be seen as a transformation of the uniform distribution $\lambda$ on $[0,1]$, via the above 'generalised inverse' $F^{-1}$. $F$ might not be injective, and it might not be surjective, but their is only ever one right-continuous choice of $F^{-1}$, which is the one stated above.
So it's not true that, on the level of maps, your probability space can always be replaced by the unit interval, but it is true that there is a sequence of $i.i.d.$ random variables on $[0,1]^{\mathbb{N}}$ (we need one copy of $[0,1]$ for each independent $Y_i$) that has the same distribution as your given sequence, no matter which probability space you started with.
I'm not sure what you mean by your last question. In your case, the $X_i$ have the same distribution, so $F$ is one, fixed map. And it is explained above that the probability of $X_i\leq x$ is the probability of $Y_i\leq F(x),$ since $F^{-1}(Y_i)$ has the same distribution as $X_i$. Again, they're not the same map.
But your statement is about every $x\in \mathbb{R}$ and hence, we can replace $F(x)$ by a general $y$ and show an, a priori, stronger statement.
