Proving that if $A_n\xrightarrow{s}A$, $\sup\limits_{n\in\mathbb N}\|A_n\|<\infty$, $B_n\xrightarrow{s}B$, then $A_nB_n\xrightarrow{s}AB$ 
Prove that if $A_n\xrightarrow{s}A$, $\sup\limits_{n\in\mathbb N}\|A_n\|<\infty$, $B_n\xrightarrow{s}B$, then $A_nB_n\xrightarrow{s}AB$.

$\xrightarrow{s}$ is the convergence wrt strong topology, $\xrightarrow{w}$ is convergence wrt weak topology. $\mathcal B(X,Y)$ is the set of all bounded function from $X$ to $Y$.
Let $X, Y, Z$ be normed spaces and $A_n, A \in \mathcal B(Y,Z),\ B_n,B \in \mathcal B(X,Y)$.
Then,
$$ \bigg(A_n \xrightarrow{s} A,\ \sup_{n \in \mathbb N} \|A_n\|<\infty,\ B_n \xrightarrow{s} B\bigg)\Rightarrow \bigg(A_nB_n\xrightarrow{s}AB\bigg)$$
$$ \bigg(A_n \xrightarrow{w} A,\ \sup_{n \in \mathbb N} \|A_n\|<\infty,\ B_n \xrightarrow{s} B\bigg)\Rightarrow \bigg(A_nB_n\xrightarrow{w}AB\bigg)$$

My question is if this also holds true if we don't have the assumption that $\sup_{n \in \mathbb N} \|A_n\|<\infty$

If $X$, $Y$, $Z$ are Banach spaces it holds true (the proof uses the principle of uniform boundedness). So I guess it does not hold true if $X$, $Y$, $Z$ are simply normed spaces (and not Banach spaces) but I have no idea how to show it
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So, by 'usual basis' you mean $e_i=(0,\ldots,0,1,0,\ldots,0)$ where we have $1$ for the i-th component and $0$ else? 
Then we have $(x_n)_{n\in \mathbb N}=(\lambda_1e_1,\ldots,\lambda_ie_i,\ldots,\lambda_Ne_N,0,0\ldots)$ for every $(x_n)_{n\in \mathbb N} \in C_{00}(\mathbb N)$, right? (because we only have finite support), $\lambda_i \in \mathbb R$ or $\mathbb C$
How does $e_n^*$ look like?
Furthermore, $A_n(e_i)=2^ne_1\otimes e_n^*(e_i)=2^n\delta_{ni}e_i=\begin{cases}2^ne_n,&i=n\\0,&i\neq n\end{cases}$
$B_n(e_i)=\frac{1}{n}e_n\otimes e_1^*(e_i)=\frac{1}{n}\delta_{1i}e_i=\begin{cases}\frac{1}{n}e_1,&i=1\\0,&i\neq 1\end{cases}$
But I don't understand how $A_n$ takes the nth component to the first component, and $B_n$ takes the first component to the nth component?
What does $A_nB_n$ do with a vector in $c_{00}(\mathbb N)$?
Basically, I don't really understand how $A_n((x_n)), B_n((x_n))$ and $A_nB_n((x_n))$ work.
 A: The statements are true generally and do not need the uniform boundedness theorem. For the first statement:
$$\|(AB-A_nB_n)x\|= \| (A-A_n)B x +A_n(B-B_n)x\|≤ \|(A-A_n)Bx\|+\|A_n\|\ \|(B-B_n)x\|$$
since $\|A_n\|$ is bounded the right-hand side goes to $0$ under your hypothesis, implying $A_nB_n\to AB$ in the strong topology. This works independently of any completeness assumptions.
For the second case do exactly the same thing, for every $f\in Z^*$ you've got:
$$|f([A_nB_m-AB]x)|≤ |f([A-A_n]Bx)| + \|f\|\ \|A_n\|\ \|(B-B_n)x\|.$$
In the case the assumption on the norms is dropped the statements are no longer true (provided the spaces are not all Banach). Consider $X=Y=Z=c_{00}(\Bbb N)$ the space of finite support sequences with supremum norm. Let $e_n$ be the usual basis and $e_n^*$ the usual dual element given by "evaluation". Then $A_n = 2^n e_1\otimes e_n^*, B_n = \frac1n e_n\otimes e_1^*$ satisfy the conditions (both converge strongly to $0$) but $A_n B_n = \frac{2^n}{n} e_1\otimes e_1^*$ doesn't converge to $0$.
(The map $e_k\otimes e_j^*$ is given by linear extension of $e_k\otimes e_j^*( e_n) = \delta_{jn} e_n$.)
