If in a Banach Space $X$ we have everywhere point wise convergent sequences of bounded operators $A_n, B_n\rightarrow A,B$ is it true that their composition is also everywhere point wise convergent?
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Yes, the composition is pointwise convergent. For any $x\in X$, one has
$$\|A_n B_n(x) - AB(x)\| \leq \|A_n B_n (x) - A_n B(x) \| + \|A_n B(x) - AB(x)\|.$$
The second term tends to zero as $n\rightarrow \infty$ since $A_n(B(x)) \rightarrow A(B(x))$. For the first term, use uniform boundedness on $(A_n)$ to get that $\alpha:= \sup\|A_n\| < \infty$. Then
$$ \|A_n B_n (x) - A_n B(x) \| \leq \alpha \|B_n(x) - B(x)\| \rightarrow 0,$$
which proves $A_n B_n (x)\rightarrow AB(x)$.