Why are generators of the Lorentz group antisymmetric, while boost matrices are symmetric? We know that a Lorentz  boost can be written as
$$
\begin{aligned}
  x_0^{\prime} &=\gamma\left(x_0-\beta x\right) \\
  x^{\prime} &=\gamma\left(x-\beta x_0\right) \\
  y^{\prime} &=y \\
  z^{\prime} &= z, 
\end{aligned}
$$
symmetric between X and t.
However, infinitesimally, it is included in
$$
\Lambda_{~~~\nu}^\mu=\delta^\mu{ }_\nu+\omega^\mu{ }_\nu,
$$
whose infinitesimal transformations amount to
$$
x^{\prime \mu}=x^\mu+\omega^\mu{ }_\nu x^\nu.
$$
Here
$$
\omega_{\mu\nu}=-\omega_{\nu\mu},
$$
antisymmetric.
Question: how is a symmetric boost transformation quantified by infinitesimal antisymmetric parameters?
 A: It's in the funny Minkowski metric. In point of fact, as a matrix, for a boost,
$$
\omega^\mu_{~~\nu} = \omega^\nu_{~~\mu},
$$
so it is symmetric, unlike the antisymmetric covariant object,
$$
\eta_{\mu\kappa} \omega^\kappa_{~~\nu} ~~~~~~~~\leadsto \\
\omega_{\mu\nu}= - \omega_{\nu\mu},
$$
as the lowering of the space-like indices pick up a sign w.r.t. the timelike index.
So, leaving the irrelevant y,z inert directions alone, your infinitesimal boost (~to lowest order in β) is but
$$
\begin{pmatrix}x^0 \\ x^1 \end{pmatrix} '= \begin{pmatrix}1&-\beta\\ -\beta & 1 \end{pmatrix}\begin{pmatrix}x^0\\ x^1 \end{pmatrix} =\left (I+ \begin{pmatrix}0& \omega^0_{~~1}\\ \omega^1_{~~0} & 0 \end{pmatrix}\right )\begin{pmatrix}x^0\\x^1  \end{pmatrix} ,
$$
since $\omega^0_{~~0}=0=\omega^1_{~~1}$.
To be sure, this mismatch miracle does not occur for rotations, which entail only spacelike indices, so the mixed tensor has the same antisymmetry as the covariant one.

*

*In conclusion, the antisymmetry of the covariant tensor $\omega_{\mu\nu}$ elegantly unifies rotations with boosts (hyperbolic rotations) by dint of the Minkowski metric. Neat, huh?


Clarification to comment question
Indeed, you don't understand the notation: The mixed tensor (one covariant and one contravariant index) is not always symmetric: only for the boost, but not for rotations. So,
for the boost,
$$
\omega_{0~1}=\eta_{0\kappa} \omega^ \kappa_{~~1}=\omega^ 0_{~~1}= \omega^ 1_{~~0}= -\omega_{1~0}\equiv b,
$$
but for a rotation,
$$
\omega_{2~1}=\eta_{2\kappa} \omega^ \kappa_{~~1}=-\omega^ 2_{~~1}= \omega^ 1_{~~2}= -\omega_{1~2}\equiv a.
$$
If we take $\omega_{0~2}=0$, and ignore the z direction, we have the mixed-symmetry mixed-tensor matrix  ,
$$
\omega^ \mu _{~~\nu} = \begin{pmatrix}0 & b&0 \\    b&0 &  a \\ 0& - a&0 \end{pmatrix},
$$
with the standard structure of the boost and rotation generators.
