You may use one of representations, then - prove the relation $\gamma_{\mu}^{+} = \gamma_{0}\gamma_{\mu}\gamma_{0}$ in this representation, and finally prove that it is correctly for an arbitrary representation (as I think that you want to see
just that).
Here I will use the spinor representation:
$$
\gamma^{\mu} = \begin{pmatrix} 0 & \sigma^{\mu} \\ \tilde {\sigma}^{\mu} & 0 \end{pmatrix}, \quad \sigma^{\mu} = (\hat {\mathbf E }, \sigma ), \quad \tilde {\sigma}^{\mu} = (\hat {\mathbf E }, -\sigma ).
$$
It's not hard to show that in this representation
$$
{\gamma^{\mu}}^{+} = \gamma^{0}\gamma^{\mu}\gamma^{0}. \qquad (1)
$$
Then let's introduce the unitary transformation $\Psi \to \hat {U} \Psi, \bar {\Psi} \to \bar {\Psi}\hat {U}^{+}$, which connects spinor and some other basis. Corresponding transformation of the gamma-matrices is $\gamma^{\mu} \to \hat {U}^{+}\gamma^{\mu}\hat {U}$. Lets see how does $(1)$ change under this transformation:
$$
{\gamma^{\mu}}^{+} \to (\hat {U}^{+} \gamma^{\mu} \hat {U})^{+} = \hat {U}^{+} {\gamma^{\mu}}^{+}\hat {U} = \hat {U}^{+} \gamma_{0}\gamma^{\mu}\gamma_{0}\hat {U} = \hat {U}^{+}\gamma_{0} \hat {U} \hat {U}^{+} \gamma^{\mu}\hat {U} \hat {U}^{+}\gamma^{0}\hat {U}=
$$
$$
=\tilde {\gamma}^{0}\tilde {\gamma}^{\mu}\tilde {\gamma}^{0}.
$$
Ok, after posting this question I have been trying to solve this and finally did, so I am posting the answer.
Just one identity is enough and that is:
$$ \gamma^\mu\gamma^\nu\gamma^\lambda = g^{\mu\nu}\gamma^\lambda + g^{\nu\lambda}\gamma^\mu - g^{\mu\lambda}\gamma^\nu + i \epsilon^{\sigma\mu\nu\lambda} \gamma_\sigma\gamma^5 $$
And a few basic properties of Dirac matrices like: $ (\gamma^5)^2 = 1 $, $ \{ \gamma^\mu, \gamma^5 \} = 0 $, $ \{\gamma^\mu, \gamma^\nu \} = 2 g^{\mu\nu} $ etc.
So we had:
$$ T = [\bar{s}\gamma^\mu\gamma^\sigma\gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu}\gamma_\mu\gamma_\sigma\gamma_\nu (1-\gamma^5) \nu] $$
We implement the above mentioned identity to the right bracket
$$ T \propto [\bar{s}\gamma^\mu\gamma^\sigma\gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu}( g_{\mu\sigma}\gamma_\nu + g_{\sigma\nu}\gamma_\mu - g_{\mu\nu}\gamma_\sigma + i \epsilon_{\alpha\mu\sigma\nu} \gamma^\alpha\gamma^5 ) (1-\gamma^5) \nu] $$
Now using using the symmetric property of the metric tensor the we simplify the expression a lot:
$$ T \propto (4+4+4) \{ [\bar{s} \gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu}\gamma_\nu(1-\gamma^5) \nu] + i \epsilon_{\alpha\mu\sigma\nu} [\bar{s}\gamma^\mu\gamma^\sigma\gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu} \gamma^\alpha\gamma^5 (1-\gamma^5) \nu] $$
Now if we again use the identity of the product of three Gamma matrices in the first bracket of the last term and again the metric tensor-terms will drop because of the anti-symmetric property of the Levi Civita and so the last term will become:
$$ T_{last} = i \epsilon^{\alpha\mu\sigma\nu} [\bar{s}\gamma^\mu\gamma^\sigma\gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu} \gamma_\alpha\gamma^5 (1-\gamma^5) \nu] = i \epsilon_{\mu\sigma\nu\alpha} i\epsilon^{\mu\sigma\nu\delta} [\bar{s}\gamma_\delta \gamma^5 (1-\gamma^5) d] \otimes [\bar{\nu} \gamma^\alpha\gamma^5 (1-\gamma^5) \nu] $$
Now we can apply the identity: $$ \epsilon_{\mu\sigma\nu\alpha} \epsilon^{\mu\sigma\nu\delta} = 4! \delta^\alpha_\delta $$ and also using the properties of $\gamma^5$ finally we have,
$$ T_{last} = - 24 [\bar{s}\gamma^\alpha(1-\gamma^5) d] \otimes [\bar{\nu} \gamma_\alpha(1-\gamma^5) \nu] $$
Hence,
$$ T \propto [\bar{s}\gamma^\alpha(1-\gamma^5) d] \otimes [\bar{\nu} \gamma_\alpha(1-\gamma^5) \nu] = (\bar{s} d)_{V-A}(\bar{\nu}\nu)_{V-A} $$
Best Answer
I figured it out with the help of the comments after some errors. $$\gamma^{\mu} \gamma^{\nu} \partial_{\mu} \partial_{\nu}= \frac{1}{2}\left(\gamma^{\mu} \gamma^{\nu} \partial_{\mu} \partial_{\nu}+\gamma^{\nu} \gamma^{\mu} \partial_{\nu} \partial_{\mu}\right)= \frac{1}{2}\left(\gamma^{\mu} \gamma^{\nu}+\gamma^{\nu} \gamma^{\mu}\right) \partial_{\mu} \partial_{\nu}=\frac{1}{2}\left\{\gamma^{\mu}, \gamma^{\nu}\right\} \partial_{\mu} \partial_{v}$$