If $\lim_{n\to \infty}a_n = a\in \mathbb{R}$ . Prove that $\limsup_{n\to \infty}a_n x_n=a\limsup_{n\to \infty}x_n$ .

For this $\lim\sup$ should not be $\infty$. Also I will assume $\lim$ and $\lim\sup$ are positive. I will do the lower bound and you should carry out the rest of the argument for the upper bound. By definition of limit and $\lim\sup$ given $\epsilon>0$ there exists $N$ such that for all $n>N$ $a_n>(1-\epsilon)a$ and for all $N$ there exists $m>N$ such that $x_m>(1-\epsilon)\lim \sup x_n$. Combining these you find $$a_mx_m>(1-\epsilon)^2 a\lim \sup x_n$$

We can always find arbitrarily large $m$ satisfying this hence $\lim\sup a_nx_n\geq (1-\epsilon)^2a \lim \sup x_n$. Now let $\epsilon\rightarrow 0$.

Let us work with the definition: $$\limsup_{n\to\infty} x_n = \lim_{n\to\infty} \sup_{k\ge n} x_k.$$ (There are several equivalent definitions of limit superior. This one seems to be very often introduced as the first of them.)

We are trying to show that if $x_n>0$, $a>0$ and $\lim\limits_{n\to\infty} a_n=a$ then $$\limsup_{n\to\infty} a_nx_n = a\limsup_{n\to\infty} x_n.$$ (It has been already pointed out in other answers and comments that the claim is not true without some assumptions on the sequences $a_n$, $x_n$; i.e., in the way it is stated in the question.)

For any given $\varepsilon>0$ we have some $n_0$ such that $k>n_0$ implies $$a-\varepsilon \le a_k \le a+\varepsilon.$$ We will only work with $\varepsilon<a$, so that we have $a-\varepsilon >0$. Then we have \begin{gather*} a-\varepsilon \le a_k \le a+\varepsilon\\ (a-\varepsilon)x_k \le a_kx_k \le (a+\varepsilon)x_k \end{gather*} for $k>n_0$ and \begin{gather*} \sup_{k\ge n} (a-\varepsilon)x_k \le \sup_{k\ge n} a_kx_k \le \sup_{k\ge n} (a+\varepsilon)x_k\\ (a-\varepsilon)\sup_{k\ge n} x_k \le \sup_{k\ge n} a_kx_k \le (a+\varepsilon)\sup_{k\ge n} x_k \end{gather*} Applying limit $n\to\infty$ to the above equation gives $$(a-\varepsilon)\limsup_{n\to\infty} x_n \le \limsup_{n\to\infty} a_nx_n \le (a+\varepsilon)\limsup_{n\to\infty} x_n.$$ Since the above inequality is true for every $\varepsilon$ such that $0<\varepsilon<a$, we get that $$\limsup_{n\to\infty} a_nx_n = a\limsup_{n\to\infty} x_n.$$

Notice that the above argument also works for $\limsup x_n=0$.

The claim from the title also works for $a=0$, but in this case we would have to disallow $\limsup x_n=+\infty$. (And the proof would be different.)

Similarly, the claim is true also in the case $\lim a_n=a=+\infty$. But now we would have to disallow $\limsup x_n=0$. (In the other words, we have to avoid indeterminate forms $0\cdot\infty$ and $\infty\cdot0$ in the product $a\cdot\limsup x_n$.)

How about $a_n = \frac{1}{n}$ and $x_n = n$?

Then $\sup a_nx_n = 1$ do we have $0 \times +\infty =1$, but then, what about $b_n = \frac{2}{n}$

Your claim that $\sup a_n x_n = \sup a_n \sup x_n$ isn't true, even for positive convergent sequences. Take (for instance) $a_n = \frac{n+1}{n}$ and $x_n = \frac{n}{n+1}$.

The proof can't be fixed because as stated the result isn't true unless the terms are nonnegative. Take $a_n = -1$ and $x_n = (-1)^n$.