Pointless characterization relating between a fractal and its code space 
Given an hyperbolic IFS $(X,\{f_i:i=1,\ldots,N\})$ and denoting its code space by $\Sigma_N = \{1,\ldots,N\}^{\mathbb{N}}$ and the generated fractal set by $\mathcal{A}$.
There is a continuous and surjective mapping $\gamma: \Sigma_N \to \mathcal{A}$ given by $\gamma(\sigma) = \lim\limits_{n \to \infty} f_{\sigma(n)}(x)$ where $x$ can be chosen arbitrarily in $X$.

Denoting for $f:X \to X$ by $\overline{f}:H(X) \to H(X)$ the function $\overline{f}(A) = \{f(a):a \in A\}$ where $H(X)$ is the hyperspace of compact subsets of $X$. The author makes the following remark:

Suppose that $\sigma \in \Sigma_N$ and let $A_{\sigma(n)} = \overline{f}_{\sigma(n)}(A)$ for $A \in H(X)$. Then the above theorem states that $\gamma(\sigma) = \bigcap\limits_{n \in \mathbb{N}} A_{\sigma(n)}$

How can I show this remark is true?
My try
$f(X) \subset X$. This is because $f:X \to X$. The strict inequality follows if one takes $x_0,y_0 \in X$ such that $diam(X) = d(x_0,y_0)$. It is clear that $x_0 \in f(X)$ and $y_0 \in f(X)$ cannot happen simultaneously, since the diameter decreases strictly:
$$diam(f(X)) = \sup\{d(x,y).x,y \in f(X)\} = \sup\{d(f(x'),f(y')).x',y' \in X\} \le \lambda \cdot \sup\{d(x',y').x',y' \in X\} = \lambda \cdot diam(X) < diam(X)$$
where $\lambda < 1$. By monotonicity, $f(f(X)) \subseteq f(X)$ and then do induction. For this decreasing sequence, one has $\cap_{i = 1}^n \{\overline{f}_{\sigma(i)}(X)\} = \overline{f}_{\sigma(n)}(X)$ so $\cap_{i = 1}^\infty \{\overline{f}_{\sigma(i)}(X)\} = \lim\limits_{n \to \infty} \overline{f}_{\sigma(n)}(X) = \lim\limits_{n \to \infty}  f_{\sigma(n)}(x)$ for $x \in X$.
But this cannot be applied to any $A \in H(X)$. It can be applied though to sets such that $f(S) \subseteq S$ like the fractal set $\mathcal{A}$.
References
These lecture notes.
Massopust's Interpolation and approximation with splines and fractals
 A: $\newcommand{\de}{\delta}
\newcommand{\ga}{\gamma}
\newcommand{\si}{\sigma}
\newcommand{\ep}{\varepsilon}$
First here, $\bigcap\limits_{n \in \mathbb{N}} A_{\sigma(n)}$ is a subset of $X$, whereas $\gamma(\sigma)$ is an element of $X$. So, in your second highlighted statement $\gamma(\sigma)$ should be replaced by $\{\gamma(\sigma)\}$. 
(This is using the notations on pages 25 and and 22 of the lecture notes: 
$$A_{\sigma(n)}:=f_{\sigma(n)}(A),\quad
f_{\sigma(n)}:=f_{\sigma_1}\circ\dots\circ f_{\sigma_n}
$$
for $\sigma=(\sigma_1,\sigma_2,\dots)\in\Sigma_N=\{1,\ldots,N\}^{\mathbb{N}}$.)
Even after that, your second highlighted statement will be definitely false if e.g. $A=\emptyset\ne X$. It will also be false if e.g. $X=[0,1]$, $f_1(x)=x/2$ for $x\in[0,1]$, $f_2,\dots,f_N$ are any contraction maps of $X=[0,1]$ into itself, $A$ is the nonempty compact set $\{1\}$, and $\si=(1,1,\dots)$. 
However, it is easy to see that 
\begin{equation*}
\{\gamma(\sigma)\} = \lim_n A_{\si(n)} 
\end{equation*}
for any nonempty bounded $A\subseteq X$, in the sense 
\begin{equation*}
 d(A_{\si(n)},\ga(\si)):=\sup\{d(y,\ga(\si))\colon y\in A_{\si(n)}\}\to0,\tag{1}
\end{equation*}
where $d$ is the distance function on $X$. 
Indeed, let $M\in[0,\infty)$ be the diameter of the bounded set $A$. Then for some $r\in[0,1)$, all $\si\in\Sigma_N$, and all natural $n$ the diameter of the set $A_{\si(n)}$ will be $\le M r^n$. 
So, for any $x\in A$
\begin{equation*}
 d(A_{\si(n)},\ga(\si))\le d(A_{\si(n)},f_{\si(n)}(x))+d(f_{\si(n)}(x),\ga(\si))
 \le M r^n+d(f_{\si(n)}(x),\ga(\si))\to0 
\end{equation*}
so that (1) follows. 
