Proving that A(B ∩ C) ⊆ AB ∩ AC
As Yuval indicated, the most powerful and most popular proof technique is to use definitions. 

The first principle of mathematical proof is by definitions. Always check the definitions. Whenever you are stuck in an unfamiliar area, look up the definitions first. We can not succeed in overemphasizing the importance of definitions. I can remember I had been chewing the definitions a dozen times in a row from time to time.

Here our task is to prove $A(B \cap C) \subseteq AB \cap AC$.

What is the definition of "$\subseteq$"? 


  If $X$ and $Y$ are sets and every member of $X$ is also a member of $Y$, then $X$ is a subset of $Y$, denoted by $X\subseteq Y$.


So according to the definition, you are supposed to show


  if $w$ is a member of $A(B \cap C)$, then $w$ is also a member of $AB \cap AC$.


What does it mean by $w$ is a member of $A(B\cap C)$? It means $w=uv$, where $u\in A$, $v\in B$ and $v\in C$ as noted correctly in the question.

What are the members of $AB\cap AC$?


  The intersection of two sets $X$ and $Y$, denoted by $X \cap Y$, is the set of all elements that are members of both the sets $X$ and $Y$. 


So you are supposed to show $w$ is a member of both $AB$ and $AC$. Since $w=uv$, where $u\in A$, $v\in B$, you know $w$ is a member of $AB$. Similarly, $w$ is a member of $AC$. Done.


  $AB \cap AC = \{ UV \mid U \in AB \text{ and } AC, V \in AB \text { and } AC \}$ for the right part.


Can you see where you made a mistake?