How to get the fourth component of EOM in a relativistic formulation of a charged particle in an electromagnetic field? We consider in Lorentz spacetime, $(x^0,x^1,x^2,x^3)=(t,x,y,z)$, choose the unit of time such that $c=1$. 
Given a four vector $A_\mu$, and let the Lagrangian 
$$L(x^i,\dot x^i,t)=-m\sqrt{1-\dot x_i\dot x^i}+qA_0+qA_i\dot x^i,\tag{1}$$
where we use Einstein's convention for summation. (See video at 35:15 with electron charge $q=-e$.)  
By the Euler-Lagrange equation
$$\frac{d}{dt}\frac{\partial L}{\partial\dot x^i}-\frac{\partial L}{\partial x^i}=0\tag{2}$$
we can get three equations, and if we use the fact that 
$$\frac{d}{d\tau}=\frac{1}{\sqrt{1-\dot x_i\dot x^i}}\frac{d}{dt}\tag{3}$$ 
then these equations are, for $i=1,2,3$
$$m\frac{d^2x^i}{d\tau^2}=q(\frac{\partial A_\mu}{\partial x^i}-\frac{\partial A_i}{\partial x^\mu})\frac{dx^\mu}{d\tau}\tag{4}$$
My question follows, can we obtain a fourth equation which is just letting $i=0$? 
The video at 59:30 says that this fourth equation is from the property that the action is invariant under Lorentz transformation. But I think this is not very convincing. I tried to view the first three equations as two four vectors having three equation space components, but I do not think this fact can lead to the equality for the last time component.
References:


*

*L. Susskind, Special Relativity, video lecture 7, May 21, 2012. 

 A: The action you start with is 
$$
S = \int d\tau L
$$
and
$$
L = - m \sqrt{ - \eta_{\mu\nu} {\dot x}^\mu {\dot x}^\nu } +q {\dot x}^\mu A_\mu \, . 
$$
This action has a gauge symmetry, which is reparameterization invariance, $\tau \to \tau'(\tau)$. In order to write down your Lagrangian you choose the gauge $\tau = x^0 = t$.
Once in this gauge, you can derive equations of motion for $x^i$ but not $x^0$ because you have gauge-fixed it. These are the ones you have written down.
However, since you are fixing a gauge, there's going to be a gauge constraint which is essentially the equation of motion derived w.r.t. $x^0$ in the first action. This is the equation you are missing. 
A: *

*Susskind's statement is the following.

If three components of a Lorentz covariant 4-vector equation $V^{\mu}=W^{\mu}$ are equal in any inertial frame, then the fourth components must be equal as well.

Proof: By linearity it is enough to consider the case where
$$ W^{\mu}~=~\begin{bmatrix} 0 \cr 0 \cr 0 \cr 0 \end{bmatrix}. $$
Now we perform an arbitrary Lorentz transformation on a 4-vector
$$ V^{\mu}~=~\begin{bmatrix} V^{0} \cr 0 \cr 0 \cr 0 \end{bmatrix}. $$
Since this does not change the three 0s by assumption, it must be because $V^0=0$ is zero to begin with. $\Box$


*Another idea to get 4 Euler-Lagrange (EL) eqs. is to have 4 independent dynamical variables $x^{\mu}$, cf. Prahar's answer. The main point is that OP's original action $S[x^1,x^2,x^3]$ can be viewed as a gauge theory $S[x^0,x^1,x^2,x^3]$ with a certain gauge-fixing condition
$$\chi~\approx~0.  \tag{A}$$
Now strictly speaking, we should introduce a Lagrange multiplier $\color{red}{B}$
to implement the gauge-fixing condition (A), i.e. include a gauge-fixing term
$$ L_{\rm gf}~:=~\color{red}{B} ~\chi \tag{B} $$
in the Lagrangian. But this raises a question.

How do we know that the presence of the Lagrange multiplier $\color{red}{B}$ does not affect the EL eqs.?

Formal Answer: Because of the (second) Noether identity, cf. e.g. Ref. 1.
Here we want to explicitly verify this fact for OP's model (1).


*We prefer to work with a non-square root$^1$ formulation$^2$
$$\begin{align} S~:=~&\int\!d\tau~L , \cr  
L~:=~&L_0 - U +L_{\rm gf}, \end{align}\tag{C}$$
with
$$\begin{align} L_0~:=~& \frac{\dot{x}^2}{2e}-\frac{e m^2}{2}, \cr 
\dot {x}^2~:=~& g_{\mu\nu} \dot {x}^{\mu} \dot {x}^{\nu}, \cr \dot {x}^{\mu} ~:=~&\frac{dx^{\mu}}{d\tau},\end{align}\tag{D}$$
and with an einbein $e>0$. Here $\tau$ is the world-line (WL) parameter (which is not necessarily proper time). The velocity-dependent Lorentz potential is
$$ U~:=~ - q{\dot x}^{\mu} A_{\mu},  \tag{E} $$
with corresponding generalized Lorentz 4-force$^3$
$$\begin{align} F_{\mu}~:=~&\frac{d}{d\tau} \frac{\partial U}{\partial \dot{x}^{\mu}} - \frac{\partial U}{\partial x^{\mu}}~=~qF_{\mu\nu}\dot {x}^{\nu}, \cr 
F_{\mu\nu}~:=~&\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}. \end{align}\tag{F}$$


*The canonical/conjugate 4-momentum$^3$ reads
$$ p_{\mu}~:=~\frac{\partial L}{\partial \dot{x}^{\mu}}~=~ p^{\rm kin}_{\mu}+ q A_{\mu}, \tag{G}$$
where the kinetic 4-momentum$^3$ reads
$$ p^{\rm kin}_{\mu} ~:=~ \frac{g_{\mu\nu}\dot{x}^{\nu}}{e}. \tag{H}$$


*Now the EL eqs. for the action (B) read
$$\begin{align} 0~\approx~&\frac{\delta S}{\delta x^{\mu}}
~=~ -\dot{p}^{\rm kin}_{\mu} +\frac{\partial_{\mu}g_{\nu\lambda}}{2e}\dot {x}^{\nu} \dot {x}^{\lambda}  +F_{\mu} + \color{red}{B}\frac{\partial \chi}{\partial x^{\mu}} \cr
~=~& -g_{\mu\nu}\frac{d}{d\tau}\left(\frac{\dot{x}^{\nu}}{e}\right) -\frac{\Gamma_{\mu,\nu\lambda}}{e}\dot {x}^{\nu} \dot {x}^{\lambda}  +F_{\mu} + \color{red}{B}\frac{\partial \chi}{\partial x^{\mu}}, \end{align} \tag{I}$$
$$ 0~\approx~\frac{\delta S}{\delta e}
~=~-\frac{1}{2}\left(\frac{\dot{x}^2}{e^2}+m^2 \right) + \color{red}{B}\frac{\partial \chi}{\partial e}, \tag{J} $$
$$ 0~\approx~\frac{\delta S}{\delta B}~=~\chi. \tag{K}$$


*The question we would like to address is the following.

How do we conclude that the Lagrange multiplier $\color{red}{B}~\approx~0$ vanishes in EL eqs. (I) & (J)?



*In anticipation of what to come, let us first differentiate eq. (J) wrt. $\tau$:
$$\begin{align} \frac{d}{d\tau}\left( \color{red}{B}\frac{\partial \chi}{\partial e}\right)
~\stackrel{(J)}{\approx}~&\frac{\dot{x}^{\mu}}{e} \left(g_{\mu\nu}\frac{d}{d\tau}\left(\frac{\dot{x}^{\nu}}{e}\right) +\frac{\Gamma_{\mu,\nu\lambda}}{e}\dot {x}^{\nu} \dot {x}^{\nu} \right)\cr 
~\stackrel{(I)}{\approx}~&\frac{\dot{x}^{\mu}}{e}\color{red}{B}\frac{\partial \chi}{\partial x^{\mu}}. \end{align}\tag{M}$$


*Static gauge $\tau=x^0$. This is OP's gauge. In the static gauge
$$ \chi~:=~x^0-\tau . \tag{N}$$
Then eq. (M) becomes
$$ 0~\stackrel{(M)+(N)}{\approx}~ \frac{\color{red}{B}}{e},   \tag{O}$$
i.e. the Lagrange multiplier $\color{red}{B}$ vanishes! $\Box$


*Affine parametrization gauge $e=1/m$.
$$ \chi~:=~e-1/m. \tag{P}$$
Then eq. (M) becomes
$$  \dot{\color{red}{B}}~\stackrel{(M)+(P)}{\approx}~0.\tag{Q} $$
We can only conclude that the Lagrange multiplier $\color{red}{B}$ is a constant. This is related to the fact that this is only a partial gauge fixing. In this gauge the world-line (WL) parameter $\tau$ is an affine function of proper time.
References:

*

*H. Motohashi, T. Suyama & K. Takahashi, Phys. Rev. D 94 (2016) 124021, arXiv:1608.00071.

--
$^1$  To achieve OP's square root formulation (1), simply integrate out the $e$ field, cf. e.g. this Phys.SE post. The gauge symmetry is e.g. discussed in this Phys.SE post and links therein.
$^2$ Notation & conventions. We use signature convention $(-,+,+,+)$ and $c=1$. The $\approx$ symbol means equality modulo eoms.
$^3$ The usual notions of 4-momentum & 4-force correspond to the gauge where the world-line (WL) parameter $\tau$ is proper time.
