Proof that $f$ is injective iff $\bigcap_{i\in I} f(A_i) = f( \bigcap_{i \in I} A_i)$ for every $(A_i)_{i\in I}$ I think I have a proof by contradiction of the following result but I am not sure about it.

Show that $f$ is injective iff $\bigcap_{i\in I} f(A_i) = f( \bigcap_{i \in I} A_i)$ for every $(A_i)_{i\in I}$.

This is my first contact with actual mathematical theory, and it has always been my biggest knowledge gap as my actual field of study(physics) relies mostly on just applying math more than really understanding it, so I can say it has been a pretty rough first contact...
My try is as follows:
Assuming f(x) is not injective, then,
$ \exists x \land x' \in X; x \neq x'; f(x) = f(x')$
Taking $x \in \bigcap_{i \in I}A_i \land x' \in A_i$, but, 
$x' \notin \bigcap_{i \in I}A_i \Rightarrow f(x) \in f(\bigcap_{i \in I}A_i) \land f(x') \notin f(\bigcap_{i \in I}A_i)$, however, $f(x) = f(x')$.
Hence, being a contradiction!
 A: In your solution, it is not clear how you can choose your $\bigcap_{i\in I} A_i$. Here are some hints; 
Hint 1: Assume that $f:X \to Y$ is not injective. Then, as you've correctly pointed out, $\exists x_1,x_2 \in X$ such that $x_1 \neq x_2$ and  $f(x_1) = f(x_2)$. Let $A_1 = \{x_1\}$ and let $A_2 = \{x_2\}$. Then $f(A_1 \cap A_2) = f(\emptyset) = \emptyset$. However, what is $f(A_1) \cap f(A_2)$? 
Hint 2: Assume that $f:X \to Y$ is injective. We want to show that $\bigcap_{i \in I} f(A_i) = f(\bigcap_{i\in I}A_i)$. 
Step 1. First we show that $\bigcap_{i \in I} f(A_i) \subseteq f(\bigcap_{i\in I}A_i)$. Let $b \in \bigcap_{i \in I} f(A_i)$. Then, we know that there is a $c \in X$ such that $f(c) = b$.  By injectivity (why?), $c \in \bigcap_{i \in I} A_i$. Then, $f(c) \in f(\bigcap_{i \in I} A_i)$ and this implies that $b \in f(\bigcap_{i \in I} A_i)$. 
Step 2. Now, we show the other inclusion. Let $b \in f(\bigcap_{i \in I}A_i)$. It's up to you to show that $b \in f(\bigcap_{i \in I} A_i)$. This is the easy direction. 
A: "Show that
 f(x) is injective $\Leftrightarrow  \bigcap_{i _\in I} (f(A_i)) = f( \bigcap_{i \in I} A_i)$"
First see that we always have the following
$$  \bigcap_{i \in I} A_i\subset A_i~~~\forall ~i \implies f( \bigcap_{i \in I} A_i)\subset \bigcap_{i _\in I} (f(A_i)) $$
so the problem reduces to 
f(x) is injective $\Leftrightarrow  \bigcap_{i _\in I} (f(A_i)) \subset f( \bigcap_{i \in I} A_i)$

Which can again be reduce to the following 
  f(x) is injective $\Leftrightarrow f(A) \cap f(B) \subset f( A\cap B).$"
  Where A and B can be replaced by $A_i$ and $A_j$ respectively.

Now let prove our last claim.

Assume f(x) is injective

Then let $z\in(A) \cap f(B) $ i.e $\{z\in f(A)  \implies \exists a\in A :f(a) = z \}$ and $\{z\in f(B)  \implies \exists b\in B :f(b) = z \}$  Thus, $f(a) = z= f(b)$. But $f$ is injective we then have $a=b \in A \cap B $ therefore $z=f(a)= f(b)\in  $z\in f(A\cap B) $. hence, $f(A) \cap f(B) \subset f( A\cap B).$

Conversely, Assume $f(A) \cap f(B) \subset f( A\cap B).$
  let prove that f is injective. Let $a$ and $b$ bet two elements in the domain of $f$ such that $f(a)=f(b)$. We set 
  $$A =\{a\}\qquad\land\qquad B=\{b\}$$
  by Asumption, we have, 

$$\{f(a)=f(b)\}=f(\{a\}) \cap f(\{b\}) \subset f( \{a\}\cap \{b\}).$$
but 
$$\{a\}\cap \{b\}= \begin{cases}\{a=b\}& if~~ a=b \\ ∅ &if ~~a\neq b\end{cases}$$
The only option we have is $a=b$ since $f(a) = f(b)\inf( \{a\}\cap \{b\}) $ that is $f( \{a\}\cap \{b\})$ is a nonempty set and we know that $f($∅$) =$∅ $\therefore f$ is an injective application.
