Why $v_i=(X_i-\mu _X)u_i$ is i.i.d? I don't understand. Ok, we have $\beta_1-\hat{\beta }_1=\frac{\frac{1}{n}\sum_{i=1}^{n}v_i}{(\frac{n-1}{n}){s_{X}}^{2}}$. So, for the first OLS assumption results that $E(v_i)=E((X_i-\bar{X})u_i\overset{p}{\rightarrow}E((X_i-\mu_X)u_i)=E(X_i-\mu_X)E(u_i)=0$ (since $E(u_i|X=x_i)=0$). Then Stock and Watson says that for the second OLS assumption $v_i$ is i.i.d. and $Var(v_i)&lt;\infty$. Surely this is for the casual sampling of variables but i can't prove it clearly. 
After this, we have $E(v_i)=0$ and $Var(v_i)={{\sigma _{v}}^{2}}&lt;\infty$ so we can apply CLT and saying that for $n\rightarrow \infty$ $v_i\sim N(0,{{\sigma _{v}}^{2}}/n)$ since $\frac{1}{n}\sum_{i=1}^{n}v_i=\bar{v}$.
Even after the comments exchange I am not totally sure what is the problem of the OP, but I will answer the question in the title.  

Assume that we have a model

$$Y_i = \beta X_i + u_i,\;\;\; i=1,...,n$$

Assume that we obtain a "random sample" of observations from $\{Y_i, X_i\}$. This makes each observation $i$ identically and independently distributed from all other observations.  

Now consider the random variable

$$v_i=(X_i-\mu _X)u_i = (X_i-\mu _X)(Y_i - \beta X_i) = X_iY_i - \beta X_i^2 - \mu_XY_i+\mu_x\beta X_i$$

while also

$$j\neq i : v_j=(X_j-\mu _X)u_j = (X_j-\mu _X)(Y_j - \beta X_j)=X_jY_j - \beta X_j^2 - \mu_XY_j+\mu_x\beta X_j$$

$v_i$ is a function of random variables $\{Y_i, X_i\}$ only, while $v_j$ is a function of random variables $\{Y_j, X_j\}$ only which are independent from  $\{Y_i, X_i\}$. So  $v_i$ is independent from $v_j$.  

More over, since $X_i$ follows the same distribution as $X_j$, and $X_j$ follows the same distribution as $Y_j$, we have that $X_iY_i$ follows the same distribution as $X_jY_j$.Etc. So we also conclude that $v_i$ will have the same distribution as $v_j$.