Both calculations are wrong, I'm afraid. Since when does $\epsilon^{\lambda\eta}\sigma_{\eta\nu} = \sigma^\lambda{}_\nu$? This is valid only for metric tensor, but $\epsilon^{\lambda\eta}$ ($\begin{pmatrix}0&1\\-1&0\end{pmatrix}$) is not a metric tensor, assuming $\epsilon^{\lambda\eta}$ is the two-dimensional Levi-Civita symbol.
Suppose $T^{\beta\delta} = \epsilon^{\alpha\beta}\sigma_{\gamma\alpha}\epsilon^{\gamma\delta}$, and the metric is identity. Since there are only 4 distinct elements, we could carry out the computation directly:
- $T^{00} = \epsilon^{10}\sigma_{11}\epsilon^{10} = \sigma_{11} = -\sigma_{00}$
- $T^{01} = \epsilon^{10}\sigma_{01}\epsilon^{01} = -\sigma_{01}$
- $T^{10} = \epsilon^{01}\sigma_{10}\epsilon^{10} = -\sigma_{10}$
- $T^{11} = \epsilon^{01}\sigma_{00}\epsilon^{01} = \sigma_{00} = -\sigma_{11}$
or use the identities $\epsilon_{ab}=\epsilon^{ab}$ and $\epsilon_{ab}\epsilon^{cd} = \delta_a^c\delta_b^d - \delta_a^d\delta_b^c$ to arrive at
$ T^{\beta\delta} = (\delta^{\alpha\gamma}\delta^{\beta\delta} - \delta^{\alpha\delta}\delta^{\beta\gamma})\sigma_{\gamma\alpha} =\delta^{\beta\delta}\operatorname{tr}(\sigma) -\sigma^{\beta\delta} = -\sigma^{\beta\delta}. $
Edit: If you define $\epsilon^{\mu\nu}$ to be the metric, then your first formula is wrong, because index raising is done by
$\huge x^{\color{red}\mu} = g^{\color{red}\mu\color{green}\nu} x_{\color{green}\nu} \tag{Correct} $
and not
$\huge x^{\color{red}\mu} = g^{\color{#FF4000}\nu\color{#808000}\mu} x_{\color{green}\nu} \tag{Wrong} $
especially when your metric $g^{\mu\nu}$ is asymmetric.