Question on minimizing the infimum distance of a point from a non compact set 
Let the non-compact $\;E\subset \mathcal X\;$ where $\;(\mathcal
 X,d_0)\;$ is a metric space and denote the distance of a point $\;x\in
 \mathcal X\;$ from $\;E\;$ as:
$\;d(x,E)=\inf_{y\in E\;}\{d_0(x,y)\}\;$.

After some research on the Internet I found that in general, if I would like to show that $\;\inf_{y\in E\;}\{d_0(x,y)\}=d_0(x,y^*)\;$ for some $\;y^* \in E\;$, then all I have to do is to take a sequence $\;y_n \in E\;$ such that $\;d(x,y_n) \to \inf_{\;y\in E\;}\{d_0(x,y)\}\;$ and find a subsequence $\;y_{n_k}\;$ of $\;y_n\;$ with $\;y_{n_k} \to y^*\in E\;$.
Why is the above sufficient? Is it related to the sequential compactness of $\;E\;$? But if it does, how is it possible since $\;E\;$ is non-compact? Did I miss some steps in the process of minimizing the infimum distance?
I would really appreciate if somebody could help me understand how this kind of exercises work and provide me (if possible) some Analysis Theorems that I might miss. 
Thanks in advance!
 A: The general approach you refer to applies to finding a solution $y^* \in E$ for any set $E$ and $x \in \mathcal X$ of the problem:
$\qquad d_0(x,y^*) = \inf_{y \in E}d_0(x,y)$
This approach relies purely on the fact that $d_0: \mathcal X^2 \to \Bbb R_{\ge 0}$ is a continuous function. Because $y_{n_k} \to y^*$, it follows that $d(x,y_{n_k})\to d(x,y^*)$.
What it does not do is establishing tot $y^*$ actually exists (even in $\mathcal X$), or is a part of $E$.
For the first, we need the metric space $\mathcal X$ to be complete (Cauchy sequences converge in $\mathcal X$). For the second, we can generally only establish $y^* \in E$ if $E$ is a closed set.
Since both these conditions were not given, we cannot ensure that this process of finding $y^*$ will actually manage to find an element of $E$ with the desired property.
A: Set $A_x\overset{_{\mathrm{def}}}{=}\{ \mathrm{d}(x,y) : y\in E\}$. By infimum definition,  we have that for all $\epsilon>0$ there is a $y_\epsilon\in E$ such that 
$$
\inf A_x\leq d(x,y_\epsilon)<\inf A_x+\epsilon
$$
Then for all $\epsilon_n=\frac{1}{n}$ there is a $y_n\in E$ such that
$$
\inf A_x\leq d(x,y_n)<\inf A_x+\frac{1}{n}
$$
The point here is that if the sequence $\{y_n\}_{n\in\mathbb{N}}$ is convergent (or has a convergent subsequence ) for a point $y^\ast\in E$  we have $\inf A_x= d(x,y^\ast)$.
Now your questions:

Why is the above sufficient? Is it related to the sequentially compactness of E?

The sequential compactness of $E$ is sufficient. 

But if it does, how is it possible since EE is non-compact?

Always some primitive notion of sequential compactness will be needed as a sufficient condition. For example, relatively compact subspace  is sucfficient. 

Did I miss some steps in the process of minimizing the infimum distance?

Having some notion of compactness, there is no loss in the constructive process of the sequence.
