Show that $S\cap(T\,\cup \,T') = (S\,\cap\,T)\cup(S\,\cap\,T')$ 
Show that $S\cap(T\,\cup \,T') = (S\,\cap\,T)\cup(S\,\cap\,T')$

I did this by creating 3 sets:


*

*$S\,\,=\{a,b,c,d\}$

*$T\,\,=\{c,d,e,f\}$

*$T'=\{a,b,c,d\}$
And then I performed the operations and checked they are the same. Is there something else I should do or only this will suffice?

Checking one example is insufficient in order to conclude a general proof.
You need to do this for abstract sets. Pick an element from $S\cap (T\cup T')$, using the definitions of $\cap,\cup$ conclude that it has to be in $(S\cap T)\cup (S\cap T')$, and then do the same thing from the other direction.
That would be a proof.
