Kinetic energy of a photon and Schwarzschild radius I have read here, that $\frac{1}{2}mv^2$ must not be applied on a photon ever. 
If i want to calculate escape velocity $v_e$ i need to use $\frac{1}{2}mv^2$ because we say that kinetic energy (positive) must be same or larger than gravitational potential (which is negative) in order for an object to escape. It is done like this:
$$
\begin{split}
W_k + W_p &= 0\\
\frac{1}{2}mv_e^2 + \left(-\frac{GMm}{r}\right) &= 0\\
v_e &= \sqrt{\frac{2GM}{r}}
\end{split}
$$
Than we say if there is a black hole inside certain radius we call Schwarzschield radius $r=R_{sch}$ not even light can escape because its escape velocity is smaller than needed to escape. At the border of the sphere with $R_{sch}$ light can barely escape, so it must hold that $c$ equals escape velocity $v_e$. So we write down the equation below and derive $R_{sch}$.
$$
\begin{split}
c &= \sqrt{\frac{2GM}{R_{sch}}}\\
R_{sch} &= \frac{2GM}{c^2}
\end{split}
$$
This is a well known equation, but it is derived allso using $\frac{1}{2}mv^2$ for light (photons). This is in contradiction with 1st statement in this post. So is the last equation even valid???
 A: The short answer is: no, $\frac{1}{2}mv^2$ is never valid for photons. A photon's energy is given by $$ E = h f = \hbar \omega = \frac{h c}{\lambda} $$
always. The derivation of the Schwarzschild radius you mention is an incorrect one that happens to give the right answer accidentally. The correct derivation requires general relativity.
A: This is one of those cases when the ends do not justify the means... Just because  you get a result that is true, from laws that aren't supposed hold in that situation, doesn't mean that the laws can be used there.
If you're asking about whether there is a deeper connection between using $\frac{1}{2}mv^2$ and getting the radius, to the best of my knowledge, there is none: it just happens to be a coincidence that they match.
Like Michael Brown said, the Schwarzschild radius can only be derived from GR, in which the Schwarzschild metric
\begin{align}
ds^2 = - (1-2GM/r) dt^2 + (1-2GM/r)^{-1}dr^2 + r^2d\Omega^2
\end{align}
is the unique maximally symmetric solution to Einstein's equations in vacuum. (I've set $c = 1$) here. Note that something interesting is happening when $r = 2GM$, and one can go on to show that $r = 2GM$ happens to be the boundary of the causal part of the future null infinity, which we identify as the event horizon.
