Showing that $E_{alpha}$ and $B_{alpha}$ is spacelike

You've got a couple of problems there. One is conceptual, and the other is notational.

Conceptually, neither $boldsymbol{E}$, nor $boldsymbol{B}$ are 4-vectors. They are the only non-zero components of the tensor $F_{alphabeta}$: $$ F_{alphabeta}=partial_{alpha}A_{beta}-partial_{beta}A_{alpha} $$ Where $A^{alpha}$ is the vector potential.

Notationally, it is convenient to distinguish space-time indices (Greek letters) from space indices in an inertial frame (Latin letters). So that, $$ E_{i}=F_{0i} $$ $$ B_{i}=-frac{1}{2}varepsilon^{0ijk}F_{jk} $$ (The minus sign can be traced down to the metric.) So that, $$ B_{3}=-frac{1}{2}varepsilon^{321}F_{21}-frac{1}{2}varepsilon^{312}F_{12}=F_{21} $$ etc. (Check me for signs, it's a good exercise.)

Also check: https://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism

You should be able to recover both Maxwell's equations, and the Lorentz force law from all this.

As to the sign of the Lagrangian (that's how it's called). Pick an inertial frame, and calculate: $$ -frac{1}{4}F_{alphabeta}F^{alphabeta}=frac{1}{2}left(boldsymbol{E}^{2}-boldsymbol{B}^{2}right) $$ Now pick particular solutions of Maxwell's equations in the vacuum (e.g. linearly polarised monochromatic waves). They have the same amplitude: $boldsymbol{E}_{0}=boldsymbol{B}_{0}$, but they are out-of-phase by half a cycle. That alone will convince you that this expression is not sign-definite: At some points it's positive, and at others it's negative. A different case is the electromagnetic energy-momentum tensor: $$ T^{alphabeta}=eta^{alphanu}F_{nugamma}F^{gammabeta}+frac{1}{4}eta^{alphabeta}F_{gammanu}F^{gammanu} $$ The (zero,zero) component giving you the energy density (frame-dependent): $$ T^{00}=frac{1}{2}left(boldsymbol{E}^{2}+boldsymbol{B}^{2}right) $$ This one is frame-dependent, but positive-definite.

Note: $c=1$ throughout.