Essentially, you want to show that each is the subset of the other, as you know.
To prove $(A\cap B)\cup C \subseteq (A \cup C)\cap B$, you start by supposing $$x \in (A \cap B)\cup C$$ and unpack what this means. The aim is to show that under this assumption, it must follow that $x \in (A \cup C)\cap B$.
So, let $x \in (A\cap B)\cup C$.
Then $x \in A\cap B$, or $x \in C$ (definition of union of sets).
This means $(x \in A$ and $x\in B$) or $x \in C$. By Demorgan's, we have $[x \in A$ or $x \in C]$ and $[(x \in B)$ or $x\in C]$. That is, $x \in (A \cup C)$ and ... etc. But there's a glitch: we can conclude $x \in (A \cup C)$,
but it does not follow that $x \in B$ or $x \in C$ implies $x \in B$. What we can show is that $$x \in (A\cup C)\cap(B \cup C)$$
And hence $$(A \cap B) \cup C \subseteq (A\cup C)\cap (B \cup C)$$
The problem is that the equivalence doesn't hold. Try to determine which inclusion (subset relation), if any, hold.* Can you find a counterexample to show the equality of the left hand side and right hand side fails to hold for all $A, B, C$?