This answer is essentially [user image's answer](https://physics.stackexchange.com/a/171121/2451) using slightly different words:

1. Argue that a Lorentz-invariant action principle for a massive point particle should be based on [proper time](https://en.wikipedia.org/wiki/Proper_time),
$$\begin{align} S[x] ~=~&\int\! d\lambda~L, \cr L~=~&\alpha \sqrt{-g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\mu}}, \cr \dot{x}^{\mu}~:=~&\frac{dx^{\mu}}{d\lambda}, \cr
g_{\mu\nu}~=~&{\rm diag}(-1,1,1,1)
\end{align}\tag{1}$$
up to a proportionality factor $\alpha$.

2. Go to static gauge $$\lambda~=~t~=~\frac{x^0}{c}.\tag{2}$$ 

3. Then the Lagrangian reads 
$$\begin{align} L~\stackrel{(1)+(2)}{=}&~\alpha\sqrt{c^2-{\bf v}^2}, 
\cr {\bf v}~:=~&\dot{\bf x}.\end{align}\tag{3}$$

4. Argue that because the Lagrangian (3) should have dimension of energy, the constant $\frac{\alpha}{m_0c}$ should be dimensionless. 

5. Argue that to recover the usual non-relativistic limit $|{\bf v}| \ll c$ for the Lagrangian $L$ [up to an irrelevant constant term that cannot be seen in the Lagrange equation, and which turns out to be the rest energy $E_0=m_0c^2$], the constant $$\frac{\alpha}{m_0c}~=~-1\tag{4}$$ must be minus one.

6. So we have derived that the Lagrangian is 
$$L~\stackrel{(3)+(4)}{=}~-m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}.\tag{5}$$

7. Therefore the canonical/conjugate momentum is 
$$\begin{align}{\bf p}~:=~&\frac{\partial L}{\partial {\bf v}}\cr
~\stackrel{(5)}{=}~&\frac{m_0{\bf v}}{\sqrt{1-\frac{{\bf v}^2}{c^2}}},\end{align}\tag{6}$$
and the energy is
$$\begin{align} E~:=~&{\bf p}\cdot {\bf v}-L\cr ~\stackrel{(5)+(6)}{=}&~m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}, \end{align}\tag{7}$$ 
which answer OP's question.

8. Furthermore, instead of just 1 point particle, we can generalize to a Lagrangian $L=\sum_{i=1}^N L_i$ for $N$ point particles, and we can even introduce interaction and potential terms. If the Lagrangian is invariant under space (explicit time) translations, then according to [Noether's theorem](https://en.wikipedia.org/wiki/Noether%27s_theorem), the total momentum (energy) is conserved, respectively.