What is the purpose of Markov chain Monte Carlo in context of simulation of distributions with hard-to-compute constants It is suggested that MCMC can be used to simulated draws from a distribution which has a complex normalising constant $C$, so that we don't need to derive them (as an example, let us suggest Beta distribution $f(x) = C x^{a-1} (1-x)^{b-1}$). We cannot calculate constant $C$ so we perform draws form a Markov chain with transition matrix provided by i.e. a Metropolis-Hastings scheme. Now, after say 1000 draws, we will have something like a histogram of visited states, which will approximate the Beta(a, b) distribution. 
BUT what is the point? We cannot simulate separate values (because states of the Markov chain are dependent, so we cannot assume that a state taken at some random time instant will be ~ Beta(a, b), but only as time -> infinity will the collective number of draws approximate the distribution in question). 
 A: 
We cannot simulate separate values (because states of the Markov chain
  are dependent, so we cannot assume that a state taken at some random
  time instant will be ~ Beta(a, b),

This is not true. Once the chain has converged, or after a "burn-in" period, every draws' marginal distributions will be the target distribution (Beta(a,b) in this case). This is because the Markov chain is stationary. If $X^i \sim \pi(x^i)$, and the transition distribution is $f(x^{i+1}|x^i)$, then at the next step $X^{i+1} \sim \int f(x^{i+1}|x^i)\pi(x^i)dx^i = \pi(x^{i+1})$. The last equality is the definition of stationarity/invariance. And yes, the draws are dependent, but there is no contradiction here.

But then we can easily depict Beta(a, b) WITHOUT the constant C,

I'm not sure if I follow this, but I will remind you that if you perform Metropolis-Hastings, at every iteration when you calculate the acceptance ratio, you do not have to evaluate the target density. Only something proportional to it. This is because the ratio of the unnormalized densities is the same as the ratio of the normalized ones, and these ratios are required for calculating the acceptance probabilities of your proposals at each iteration.

calculate numerically its integral, then take a number of points, say,
  100,000, and divide this quantity into separate bins that we would
  have formed by dividing the random variable support into equal bin
  ranges, and assigning each bin a share of points out of 100,000
  (depending on the area of Beta(a, b) over the bin range as
  percentage).

When you "calculate numerically" an integral, you're taking a sample average using all of your samples. Say you are interested in estimating the mean of your Beta(a,b) distribution. Then you take 
$$
\frac{1}{10000}\sum_i x^i,
$$
and this approximates the integral/expectation. The reason this works is because the Markov chain designed by the Metropolis-Hastings algorithm is irreducible, and has as its stationary distribution the thing you're interested in.
The separation of the support into bins is only done for plotting purposes.

Both methods do not allow us to generate 1 separate draw from the
  distribution in question (or there is a way?).

I don't think you are describing two techniques. You are referring to one technique, the MH algorithm, and describing two things that you may do with your samples (as far as I can tell).
A: In complement to Taylor's fine answer, let me signal that a Markov chain that is not strongly irreducible, that is, a chain that can only move to a limited number or range of values as in the "3 is followed by 2 or 4" example in the comments, can be turned into one by random (e.g., Poisson) subsampling: 
Given a transition kernel or matrix K, at iteration $t$, generate $X_{t+1}$ as$$Y \sim K^\kappa(X_{t},\cdot),\qquad\kappa\sim\mathcal{P}(\lambda),\quad\lambda>0$$ 
as this makes the Markov chain strongly irreducible, i.e., able to go from anywhere to anywhere.(The associated Markov kernel is called the skeleton in Meyn and Tweedie (1993).)
