MC Sampling from a Posterior Distribution Bayes rule is: 
$p(x|D)\propto p(D|x)p(x)$
Imagine $p(x)$ and $p(D|x)$ are both Gaussian distributed. I understand in this case that I can obtain the posterior analytically (conjugate priors). But I'm a bit confused on how to sample from the posterior distribution via an MC (Monte Carlo) method. 
If $p(x)$ is Gaussian, I can easily pull a sample from there, but then do I put this sample into the likelihood function? If that's the case I'm evaluating a probability value at a particular point, and not actually "taking a sample". I'm a little confused on this front. Basically once I take a sample from $p(x)$ what is the procedure I need to do with that sample (i.e. where / how do I pass it on). 
I would just like to see a histogram of this posterior distribution. 
 A: You worte 

If $p(X)$ is Gaussian, I can easily pull a sample from there, but then do I
  put this sample into the likelihood function? If that's the case I'm
  evaluating a probability value at a particular point, and not actually
  "taking a sample

I think you misunderstand the concept of conditional probability, when we sample from the prior and plug the sampled value into the likelihood , we do not evaluate a probability value at a particular point we just get a conditional distribution given the sampled value from the prior. Suppose that the target distribution is the joint probability distribution of
two variables, a discrete variable $\delta \in \{1,2,3\}$ and a continuous variable $\theta \in \mathbb{R}$ then the target density  will be defined as :
01-The discrete part of this density is : 
$$ \{Pr(\delta =1), Pr(\delta = 2), Pr( \delta= 3)\} = (.45, .10, .45)$$ 
02- The continuous part of this density is 
$$P(\theta|\delta) = dnorm(\theta,\mu_{\delta},\sigma_{\delta })$$, 
where $(\mu_1,\mu_2,\mu_3) = (−3, 0, 3)$ and $(\sigma^2_1,\sigma^2_2,\sigma^2_3) = (1/3, 1/3, 1/3) $.
This is a mixture of three normal densities and a plot of the exact marginal density of ,$p(\theta) =\sum p(\theta|\delta)p(\delta)$ will be like that
 
that can  be done using the following code in R :`
mu<-c(-3,0,3)
s2<-c(.33,.33,.33)
w<-c(.45,.1,.45)

ths<-seq(-5,5,length=100)
plot(ths, w[1]*dnorm(ths,mu[1],sqrt(s2[1])) +
       w[2]*dnorm(ths,mu[2],sqrt(s2[2])) +
       w[3]*dnorm(ths,mu[3],sqrt(s2[3])) ,type="l" )

Now it is very easy to obtain independent Monte Carlo samples from the joint
distribution of $(\delta,\theta)$ First, a value of $\delta$ is sampled from its marginal distribution, then the value is plugged into $p(\theta|\delta)$ from which a value of $\theta$ is sampled (Note the sampled pair $(\delta,\theta)$ represents a sample from the joint distribution $p(\delta,\theta)=p(\delta)p(\theta|\delta)$) then the empirical distribution of the $\theta$ samples  provides an approximation to the marginal distribution $p(\theta) =\sum p(\theta|\delta)p(\delta)$ . Now a histogram of 1,000 Monte Carlo -values generated in this way or in other words the empirical distribution of the Monte Carlo samples looks like $p(\theta)$.

set.seed(1)
S<-2000
d<-sample(1:3,S, prob=w,replace=TRUE)
th<-rnorm(S,mu[d],sqrt(s2[d]))
THD.MC<-cbind(th,d)

ths<-seq(-6,6,length=1000)
plot(ths, w[1]*dnorm(ths,mu[1],sqrt(s2[1])) +
       w[2]*dnorm(ths,mu[2],sqrt(s2[2])) +
       w[3]*dnorm(ths,mu[3],sqrt(s2[3])) ,type="l" , xlab=expression(theta),ylab=
       expression( paste( italic("p("),theta,")",sep="") ),lwd=2 ,ylim=c(0,.40))
hist(THD.MC[,1],add=TRUE,prob=TRUE,nclass=20,col="gray")
lines( ths, w[1]*dnorm(ths,mu[1],sqrt(s2[1])) +
         w[2]*dnorm(ths,mu[2],sqrt(s2[2])) +
         w[3]*dnorm(ths,mu[3],sqrt(s2[3])),lwd=2 )

For more details see A First Course in Bayesian Statistical Methods the book provides also a very good comparison between MC and MCMC Chapter 6 section "Introduction to MCMC diagnostics ".
`
A: Note that you have a sample, i.e. the data, and you want to find the credibility of an unknown parameter, below I explain a special kind of MC nemely MCMC, markov Chain Monte Carlo.  
Maybe it's good to make the notation a bit more explicit, and for simplicity take the one-dimensional case (this is only for simplicity, it also works in the multi-dimensional case).  You say that your likelihood is Gaussian just as well as your prior.  If we take the one-dimensional case then our goal it to estimate one parameter, e.g. the unkown mean $\mu$.  
The posterior is then $P(\mu|_D) \propto P(D|_\mu) p(\mu)$
Now in MCMC (Monte Carlo Markov Chain), you start from an initial value $\mu=x_1$ and make a proposal move $x_2$, in order to decide whether the proposal will be accepted, you should evaluate the posterior at $x_1$ and at $x_2$. 
So you have to compute $ P(D|_\mu=x_1) p(\mu=x_1)$ (and the same for $x_2$). The prior is known (you have chosen it yourself), so you can evaluate it at $x_1$.  
The likelihood $P(D|_\mu=x_1)$ at $x_1$ is also knwown; you know it is gaussian and (in the one dimensional case) only the mean is unknown but you say that you want to evaluate it at $\mu=x_1$ so you have the mean and you can compute the value for $P(D|_\mu=x_1)$ (the data $D$ is also known of course).  Similar for $X_2$. So it is the 'Gaussian formula'' where you plug in $\mu=x_1$ and $X=D$ ($\sigma$ is known because we assumed the one dimensional case for simplicity, so only one parameter, $\mu$ is unknown, but you evaluate at $\mu=x_1$, so that's solved). 
So we can compute (up to a constant) $\pi_1=P(\mu=x_1|_D)$ and $\pi_2=P(\mu=x_2|_D)$. 
Remember that $x_2$ was only a proposal move.  This proposal is accepted if $\pi_2$ > $\pi_1$ , if $\pi_2 \le \pi_1$ then the move is accepted with a probablity $\frac{\pi_2}{\pi_1}$ (P.S. a symmetric transition kernel is assumed).
If the move is accepted then move to $x_2$ else stay at $x_1$ and next generate the next proposal, ...
This algorithm is constructed such that the so-called ''balance equations'' are fulfilled, and that is suffcient for the chain to converge to the posterior ''in the end''. 

Summary: if you can't sample from the posterior directly, then construct a markov chain that has the posterior as limiting distribution and use this Markov chain to ' walk to' that limiting distribution. The algorithm supra defines such a markov chain using a special case of the Metropolis-Hastings procedure.

[[Note, if your data is a sample with multiple elements, then you take a product of likelihoods, assuming independence). ]]
A: I agree with Der Fluß's answer and just want to add some comments. 
I think the key point here is:
In order to get samples from p(D|x)p(x), you cannot sample from p(x) and hope p(D|x) magically make the samples look like sampled from p(D|x)p(x). You have to directly sample from p(D|x)p(x).
