functions - $f(A cap B)subset f(A)cap f(B)$, and otherwise?

I got a serious doubt ahead the question

Be $f:X\longrightarrow Y$ a function. If $A,B\subset X$, show that $f(A \cap B)\subset f(A)\cap f(B)$

I did as follows

$$\forall\;y\in f(A\cap B)\Longrightarrow \exists x\in A\cap B, \text{ such that } f(x)=y\\ \Longrightarrow x \in A\text{ and }x\in B\Longrightarrow f(x)\in f(A)\text{ and }f(x)\in f(B)\\ \Longrightarrow f(x)\in f(A)\cap f(B)\Longrightarrow y\in f(A)\cap f(B)$$

This ensures that $\forall y \in f(A\cap B)$ then $y\in f(A)\cap f(B)$, therefore $f(A\cap B)\subset f(A)\cap f(B)$.

Okay, we have the full demonstration.

We know that for equality to be valid, then $f$ must be injective. But my question is when should I see that equality is not worth, not by counter example, but finding an error in the following demonstration

$$\forall\;y\in f(A)\cap f(B)\Longrightarrow y\in f(A)\text{ and }y\in f(B) \Longrightarrow \\ \exists x\in A \text{ and } B, \text{ such that } f(x)=y\\ \Longrightarrow x \in A\cap B\ \Longrightarrow f(x)\in f(A\cap B)\Longrightarrow y\in f(A\cap B)$$

Where is the error in the statement? Which of these steps can not do and why?

Answer

you wrote :

$$\forall\;y\in f(A)\cap f(B)\Longrightarrow y\in f(A)\text{ and }y\in f(B) \Longrightarrow \\ \exists x\in A \text{ and } B, \text{ such that } f(x)=y\\ $$

The problem is in the last implication : from $y\in f(A)\text{ and }y\in f(B)$ you get that there exist $x_A\in A$ and $x_B\in B$ such that $f(x_A)=y=f(x_B)$, you cannot assume that $x_A=x=x_B$.