Identity Involving Grassmann Variables and Pauli Matrices 
I am trying to prove the following identity:
  $$\theta\sigma^{\mu}\bar{\theta}\theta\sigma^{\nu}\bar{\theta}=\frac{1}{2}g^{\mu\nu}\theta\theta\bar{\theta}\bar{\theta}$$
  Where $\theta$ and $\bar{\theta}$ are anticommuting grassmann variables, and $g^{\mu\nu}$ is the metric.

I have written the above in component notation;
$$\theta^{\alpha}\sigma_{\alpha\dot{\alpha}}^{\mu}\bar{\theta^{\dot{\alpha}}}\theta^{\beta}\sigma_{\beta\dot{\beta}}^{\nu}\bar{\theta^{\dot{\beta}}}$$
Also, given the identity: $\sigma^{\mu}_{\alpha\dot{\alpha}}\sigma_{\mu\beta\dot{\beta}}=2\epsilon_{\alpha\beta}\epsilon_{\dot{\alpha}\dot{\beta}}$, I have managed to show that;
$$
\sigma^{\mu}_{\alpha\dot{\alpha}}\sigma_{\beta\dot{\beta}}^{\nu}
=
\frac{1}{2}\epsilon_{\alpha\beta}\epsilon_{\dot{\alpha}\dot{\beta}}g^{\mu\nu}
$$
Which seems like a step in the right direction.
Also, I can't seem to figure out if the barred and unbarred grassman variables commute, i.e. do we have that $\theta^{\alpha}\bar{\theta^{\dot{\beta}}}=-\bar{\theta^{\dot{\beta}}}\theta^{\alpha}$?
I have been stuck for quite a while with no useful progress, but will update if I make any. A push in the right direction would be much appreciated.
 A: The anti commuting grassman variables satisfy 
$$
\{ \theta^\alpha , \bar{\theta}_{\dot{\gamma}}\} = 0.
$$
If your concern is about the position of the indices just multiply both sides by $\epsilon^{\dot{\beta}\dot{\gamma}}$ and you obtain 
$$
\{ \theta^\alpha , \bar{\theta}^{\dot{\beta}}\} = 0.
$$
Now, as for your main question I'll go ahead and try the proof.

First note that you have a minus sign missing in your stated identity. The correct identity is 
$$
\sigma^{\mu}_{\alpha\dot{\alpha}}\sigma_{\beta\dot{\beta}}^{\nu}
=
-\frac{1}{2}\epsilon_{\alpha\beta}\epsilon_{\dot{\alpha}\dot{\beta}}g^{\mu\nu}
$$
as stated here in the very colorful fun with spinor indices

Proof: We wish to show that 
$$ \theta\sigma^{\mu}\bar{\theta}\theta\sigma^{\nu}\bar{\theta}=\frac{1}{2}g^{\mu\nu}\theta\theta\bar{\theta}\bar{\theta}.$$
Writing the LHS in index notation we have that
\begin{align*}
&\theta\sigma^{\mu}\bar{\theta}\theta\sigma^{\nu}\bar{\theta} = \theta_\alpha\sigma^{\mu}_{\alpha\dot{\alpha}}\bar{\theta}_{\dot{\alpha}} \theta_\beta\sigma^{\mu}_{\beta\dot{\beta}}\bar{\theta}_{\dot{\beta}}\\
&= \theta_\alpha\bar{\theta}_{\dot{\alpha}} \theta_\beta\bar{\theta}_{\dot{\beta}}  \sigma^{\mu}_{\alpha\dot{\alpha}}\sigma^{\mu}_{\beta\dot{\beta}}\\
&= - \frac{1}{2} \theta_\alpha\bar{\theta}_{\dot{\alpha}} \theta_\beta\bar{\theta}_{\dot{\beta}}  \epsilon_{\alpha\beta}\epsilon_{\dot{\alpha}\dot{\beta}}g^{\mu\nu} \tag{a}\\
&=  \frac{1}{2} (\epsilon_{\alpha\beta}\theta_\alpha\theta_\beta)(\epsilon_{\dot{\alpha}\dot{\beta}}\bar{\theta}_{\dot{\alpha}} \bar{\theta}_{\dot{\beta}})  g^{\mu\nu} \tag{b}\\
&= \frac{1}{2}g^{\mu\nu}\theta\theta\bar{\theta}\bar{\theta}
\end{align*}
where to get from (a) to be we have anti-commuted through the innermost grassman variables.
