Why is $f(A \cup B) = f(A) \cup f(B)$ but $f(A \cap B) \subseteq f(A) \cap f(B)$? Why is (1) $f(A \cup B) = f(A) \cup f(B)$ but (2) $f(A \cap B) \subseteq  f(A) \cap f(B)$ ? (in the case of a linear transformation $f : X \rightarrow Y$, where $A, B \subset X$)
My sheet proves that the first one is right that way :
$f(A \cup B) = \{f(x) : x \in A \cup B\}$
$f(A \cup B) = \{f(x) : x \in A \vee x \in B\}$
$f(A \cup B) = \{f(x) : x \in A\} \vee \{f(x) : x \in B\} = f(A) \vee f(B) = f(A) \cup f(B)$
But I don't understand why using a similar method won't work in the second case :
$f(A \cap B) = \{f(x) : x \in A \cap B\}$
$f(A \cap B) = \{f(x) : x \in A \land x \in B\}$
$f(A \cap B) = \{f(x) : x \in A\} \land \{f(x) : x \in B\} = f(A) \land f(B) = f(A) \cap f(B)$
It seems there is a concept I'm missing here.
 A: The way to see this is by example. For instance, what happens if $f$ is a constant function but $A$ and $B$ are disjoint? 
As for where the logic fails, it is in your third step. What you've done is you've turned an intersection of the $x$'s into an intersection of the ranges. You've broadened the sets you're intersecting by doing so (which explains the counter example above). 
Here is something to consider. My example dealt with the case where their intersection was empty. If $A\cap B$ is nonempty, is it true? 
A: If $x\neq y$ and $f(x)=z=f(y)$.
Then $z\in f(\{x\})\cap f(\{y\})$ and $f(\{x\}\cap\{y\})=f(\varnothing)=\varnothing$.
A: The problem with your book is that it abuses logical connectives. For example, $f(A \cup B) \iff \{f(x) : x \in A \vee x \in B\}$ makes no sense because $f(A \cup B)$ is not a statement, it is just a plain old set. That alone should have alerted you when repeating the book's "reasoning".
Back to the main problem. The problem with your "proof" is that $f(A) \cap f(B) \subseteq f(A \cap B)$ does not hold in general. Here is why (including an example of how the book and you should be reasoning): Say $y \in f(A) \cap f(B)$. Then $y \in f(A)$ and $y \in f(B)$. Uptil now, everything is fine. Now here's the twister: then, there is some $x \in A$ and some $x' \in B$ s.t. $y = f(x) = f(x')$. Unfortunately, you don't know whether $x = x'$! If they were, you would indeed have that $x = x' \in A \cap B$ so that $y \in f(A \cap B)$. But, as you can see from other people's answers, this is not true in general. As long as $A$ and $B$ are different sets, you can have two different $x$'s from each going to the same $y$ inside $f(A) \cap f(B)$.
A: Hint: We know $a\cap b \subseteq a$ and $a \cap b \subseteq b$. Hence $f(a\cap b) \subseteq f(a)$ and $f(a\cap b)\subseteq f(b)$ whence $f(a\cap b)\subseteq f(a)\cap f(b)$. Can you take it from here.
A: Let $A$ be the square $[0,1]\times[0,1]$ and let $B$ be $[0,1]\times[2,3]$. That is, let $A$ and $B$ be two squares on top of each other. Let $f$ be the projection onto the $x$-axis, mapping $(x,y)\mapsto x$. Then $f(A\cap B)=\varnothing$ and $f(A)\cap f(B)=[0,1]$.
Changing $A$ and $B$ can generate examples in which neither $f(A\cap B)$ nor $f(A)\cap f(B)$ are empty.
