$(\liminf_{n}A_{n})\cap (\limsup_{n}B_{n}) \subset \limsup_{n}(A_{n}\cap B_{n})$

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$(A_{n})_{n}$ and $(B_{n})_{n}$ sequences of subsets of $Ω$. Show that:

$$(\liminf A_{n})\cap (\limsup B_{n}) \subset (\limsup A_{n})\cap (\limsup B_{n})$$ and $$(\liminf A_{n})\cap (\limsup B_{n}) \subset \limsup (A_{n}\cap B_{n})$$

I am a bit confused with the set theory but I tried to prove them. I know that $\liminf A_{n} \subset \limsup A_{n}$ so the first part looks easy. For the second part I could use the first one and then I could show that $(\limsup A_{n})\cap (\limsup B_{n}) \subset \limsup (A_{n} \cap B_{n})$ but how can I know if this is always true?

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$x\in\lim\inf A_n$ means that $x$ belongs to almost all (i.e. all but finitely many) of the sets $A_n$.

$x\in\lim\sup B_n$ means that $x$ belongs to infinitely many of the sets $B_n$.

If $x$ belongs to both, it then means that $x$ also belongs to infinitely many of the sets $A_n\cap B_n$ (precisely the infinitely many sets where $x\in B_n$, minus finitely many of those where $x\not\in A_n$).

This proves that $x\in\lim\sup (A_n\cap B_n)$.