Why not any random variable is an unbiased estimator of the population mean? To prove sample mean is an unbiased estimator of the population mean, we use the following step, 
$\mathbb{E}[\sum \frac{X_i}{n}] = \frac{1}{n} \sum \mathbb{E}[X_i] = \mu$, given that $\mathbb{E}[X_i] = \mu$. 
Now my question is if we take say first random variable $X_1$ as an estimate we can go ahead and say that since $\mathbb{E}{[X_1]} = \mu$, $X_1$ is an unbiased estimator. Does that mean $X_1$ or for that matter any value is an unbiased estimator? 
I know it is not. I am doing a conceptual mistake in thinking, please let me know where I am going completely wrong.
[Edit]
As commented below for iid's every sample is indeed an unbiased estimate of the mean.
 A: It looks like you are making the mistake of confusing data with the random variables. 
Let a random variable $X$ have mean $\mu$. Then if $(x_1, \ldots x_n)$ are realizations from $X$, $\frac{1}{n} \sum x_i = \hat{\mu}$ is an unbiased estimator of $\mu$.  This is because $\mathbb{E}[\mu - \hat{\mu}] = 0$, a fact that we will use later. In fact if the data is drawn IID from $X$ then each individual sample is an unbiased estimator of the mean; the reason we take many samples is to lower the variance of our estimate.
Now if, as you mentioned, we have multiple random variables, $X_1, \ldots, X_p$, each of which has mean $\mu$, then $\mathbb{E}\left[\frac{1}{p}\sum_{i=1}^p X_i \right] = \mu$ as you suggest. Furthermore,  $\mathbb{E}X_1 = \mu$. That is, $\mathbb{E}[\mu - \mathbb{E}X_1]=0$. This implies that the observed realization of $X_1$ is an unbiased estimator of $\mu$.  There is nothing startling here but it is distinct, I think, from the question you meant to be asking which was answered in the previous paragraph. This distinction is important and worth understanding well.
