prove: max (w(E), w(E)) is a 1/2 approximation to the value OPT

I don’t know the definition of $1/2$-approximation algorithm, but it looks like what you will need to do is the following argument:

For an enumeration $sigma:Vto{1,2,…,n}$ of the vertices, let, for each input graph $G=(V,E)$, the output of the algorithm be $A_sigma(G)=max(w(overrightarrow{E}),w(overleftarrow{E}))$.

We have that $OPT(G)leq w(E)$.

On the other hand,
$$
begin{align}
A_sigma(G)&=max(w(overrightarrow{E}),w(overleftarrow{E}))\
&geqfrac{w(overrightarrow{E})+w(overleftarrow{E})}{2}\
&= frac{w(E)}{2}
end{align}$$

The last equation is because $overrightarrow{E},overleftarrow{E}$ is a partition of $E$.

Putting these two inequalities together we get $$OPT(G)leq w(E)leq 2A_sigma(G)$$

Therefore $$frac{A_sigma(G)}{OPT(G)}geq frac{1}{2}$$