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?
 A: 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.
A: A different approach: let $x$ be any element in the universe. The following equivalences hold:
$$\begin{align} x\in S\cap (T\cup T') &\iff x\in S\land x\in T\cup T'\\
&\iff x\in S\land (x\in T\lor x\in T')\\
&\iff (x\in S\land x\in T)\lor (x\in S\land x\in T')\\
&\iff x\in S\cap T\lor x\in S\cap T'\\
&\iff x\in (S\cap T)\cup (S\cap T').\end{align}$$
A: Let $X$ be a set. For every $A,B,C$ subsets of $X$, you can check that
$$\frac{(A\cap B)\subseteq C}{B\subseteq (A\rightarrow C)}$$
where $A\rightarrow C$ is defined as $(A\rightarrow C):=(\mathscr{C}A)\cup C$ and $\mathscr{C}A$ is the complementary set of $A$ in $X$ and horizontal line means "if and only if", i.e. the line above is true if and only if the line below is true.
More concisely,
$$A\cap (-)\dashv A\rightarrow (-)$$
where $\dashv$ means left adjoint, $A\cap(-)$ is the function sending a subset $B$ of $X$ to $A\cap B$ and $A\rightarrow (-)$ the function sending a subset $C$ of $X$ to $A\rightarrow C$.
Being a left adjoint, $A\cap (-)$ preservs colimits, in particular it preservs unions, hence your desidered distributive law.
$\textrm{EDIT}$: let $f:X\longrightarrow Y$ be a function, $X,Y$ sets, $A$ subset of $X$, $B$ subset of $Y$. Then you have
$$\frac{f(A)\subset B}{A\subset f^{-1}(B)}$$
Proceeding as in the main part of this answer, you have that the direct image of $f$ is left-adjoint to the inverse image $f^{-1}$, thus, being a left adjoint, direct image preserves colimits, for example unions, so that you immediately prove that $f(A\cup C)=f(A)\cup f(C)$, without computing on elements ( pick an element in $A\cup C$, do the image under $f$, check this image is in the image of $A$ or in the image of $C$, conversely, pick an element which is in $f(A)$ or in $f(B)$, check that it comes from an element....), while inverse image $f^{-1}$, being a right-adjoint, preserve limits, for example intersections, so that you have an "automatic" proof that $f^{-1}(A\cap C)=f^{-1}(A)\cap f^{-1}(C)$
So two facts apparently different as your distributive law and these properties of $f$ and $f^{-1}$ appear in some sense as two different manifestations of the same phenomenon
A: HINT: is it possible to write the left hand side otherwise? Maybe draw it first, so you have a good idea.
