Why do we sample from the uniform distribution in Metropolis-Hastings for acceptance? For each iteration of the MH, sample $x'=q(x|x')$, then the acceptance probability is computed:$$A=\min(1,a)$$
where
$$
\alpha=\frac{p(x')q(x|x')}{p(x)q(x'|x)}
$$
Now, I've seen that the algorithm samples from a uniform distribution $$
u\sim U(0,1)
$$to set the new sample as$$
x=x'
$$if $u≤A$, retain $x=x$ otherwise.

I have seen some derivation on how the MH algorithm samples from the target distribution $p$ as the stationary distribution of the Markov chain:
$$
p(x'|x)=q(x'|x)A
$$ if $x'≠x$, otherwise$$
p(x'|x)=q(x|x)+\sum_{x≠x'}q(x'|x)A
$$
Now what does the sampling from the uniform distribution have to do with deciding whether to accept the proposal $x'$ or retain $x$?
Isn't it sufficient to accept the proposal if $A>0.5$?
 A: The (simplest) validation of the Metropolis-(Rosenbluth-)Hastings algorithm is that the associated Markov kernel $K$ satisfies the so-called detailed balance equation
$$\forall\ x,x^\prime,\quad p(x)K(x,x^\prime)=p(x^\prime)K(x^\prime,x)\tag{1}$$
since (1) implies that $p(\cdot)$ is stationary. This Markov kernel measure writes as the mixture
$$K(x,\text dx^\prime)=q(x^\prime|x)\alpha(x,x^\prime)\,\text dx+\int\{1-\alpha(x,y)\}q(y|x)\,\text dy\,\delta_x(dx^\prime)\tag{2}$$
where $\delta_x(dx^\prime)$ denotes the Dirac measure at $x$. This representation means that the acceptance probability $\alpha(x,x^\prime)$ is crucial for the stationarity property (1) and cannot be replaced with
$$\displaystyle{\mathbb I_{\displaystyle\alpha(x,x^\prime)>0.5}}$$
which would lead to another stationary distribution, as can be checked on a simple example.
Here is for instance a Binomial target $\mathcal B(.1)$ and a Uniform $\mathcal U(\{0,1\})$ proposal, where the
$$\displaystyle{\mathbb I_{\displaystyle p(x^\prime)>0.5p(x)}}$$
acceptance step is failing to recover the target:
p=function(x)ifelse(x,.1,.9)
x=y=sample(0:1,N<-1e3,rep=TRUE)
for(t in 2:N)x[t]=ifelse(p(y[t])>.5*p(x[t-1]),y[t],x[t-1])

since it returns a point mass at zero instead:
summary(x)
Min.   Median  Mean   Max. 
0.000  0.000   0.001  1.000

the explanation being that $0$ is a fixed (or cemetery) state, $1$ being transient.
In practice, implementing a simulation from the kernel $K$ means

*

*selecting between the Dirac (reject) and the non-Dirac (accept) parts of the kernel in (2), and then

*simulating from the selected part (which is immediate when the Dirac part is selected).

The non-Dirac part is selected with probability
$$\int \alpha(x,x^\prime)\, q(x^\prime|x)\,\text dx\tag{3}$$
whose unbiased estimator is
$$\mathbb I_{\displaystyle U<\alpha(x,X^\prime)}\qquad \text{when}\ U\sim\mathcal U(0,1)\,,\ X^\prime\sim q(x^\prime|x)$$
Hence simulating the Uniform variate $U$ is a practical (and standard) way to achieve the realisation of an event with probability (3). In simulation terms, $U$ is called an auxiliary variable in the sense that it is not connected with the original problem of generating a Markov chain from the kernel $K$.
Note that the above explanation inverts the usual steps of a Metropolis-(Rosenbluth-)Hastings algorithm, where $X^\prime$ is first generated, then used to decide between the Dirac (reject) and the non-Dirac (accept) parts of the kernel. This double use of $X^\prime$ is both correct and more efficient than generating an independent $X^\prime$ from
$$q(x^\prime|x)\alpha(x,x^\prime)\Big/ \int \alpha(x,y)\, q(y|x)\,\text dy$$
which most often multiple rejections.
