Did I do this Set Theory proof right? so i was asked to prove $z \cap(s \cup t ) = ( z \cap s ) \cup (z \cap t)$
my proof was as follows:
set set $z \cap(s \cup t ) = ( z \cap s ) \cup (z \cap t)$ must contain the set $z$
since $ z \subset  (z \cap  t )$ and $ z \subset  (z \cap  s )$
the set must contain $s \cup t $ beacuse the set is an intersection of two sets both containing s and t
since the set contains $s \cup t $ and $z$ the set can be written $z \cap(s \cup t )$
please tell me if i did this right / what i could do better
 A: 
set $z \cap(s \cup t ) = ( z \cap s ) \cup (z \cap t)$ must contain ...

Here you are assuming they are equal, although that's the very thing you're trying to prove. That is a logical fallacy.

$z \subset  (z \cap  t )$ and $z \subset  (z \cap  t )$

That is false. Both of them. Here the inclusion is in the opposite direction: $z\supset z\cap t$ and $z\supset z\cap s.$

the set must contain ...

Which set must contain whatever it is you're saying it must contain? Do you mean $z \cap(s \cup t )$ must contain something? Or $( z \cap s ) \cup (z \cap t)$ must contain something?
That those two are equal is something that you do not know until you have proved it.

since the set contains $s \cup t $ and $z$ the set can be written $z \cap(s \cup t )$

But the set $\{1,2,3,4,5,6\}$ "contains" (I prefer to say "includes") the set $\{1,2\}$ and the set $\{5,6\}$ but nonetheless the set $\{1,2,3,4,5,6\}$ cannot be written as $z\cap(\{1,2\}\cap\{5,6\},$ no matter what set $z$ is.
If you want to prove $z \cap(s \cup t ) = ( z \cap s ) \cup (z \cap t),$ you should attempt to prove that every member of $z \cap(s \cup t )$ is a member of $( z \cap s ) \cup (z \cap t)$ and also that every member of $( z \cap s ) \cup (z \cap t)$ is a member of $z \cap(s \cup t ).$
