Before stating my questions, let me first state my understandings. Note this post is talking on metric spaces case. For the definition of a set $E$ being open in a metric space $(X,d)$, we mean that for every $a\in E$, there exists a ball $B(a,r)=\{x\in X|d(x,a)<r\}$ such that $B(a,r)\subseteq E$. Notice that the ball in the definition is with respect to $(X,d)$ (the ball centered at point $a$ and consist of all point in $X$ such that $d(x,a)<r$), it is quite important. Since every subset of a metric space is always a metric subspace; and every metric subspace is open in itself, by trivial reason. If we consider $E\subseteq X$ to be a metric space, then $E$ is always open in $(E,d)$.
For example, $[0,1]$ is not open in $(\mathbb{R},d)$; however, $[0,1]$ is open in $([0,1],d)$. So what metric space the set $E$ is with respect to play an important role when talking whether the set $E$ is open.
Now go to the definition of compactness, and also sequentially compact. It has two variations the definition would make:
(For $\textbf{compactness}$)
Let $(X,d)$ be a metric space, let $E\subseteq X$, if every open covering $\mathcal{A}$ of $E$ contains a finite subcollection that also covers $E$, then $E$ is called a $\textbf{compact set}$.
Let $(X,d)$ be a metric space. If every open covering $\mathcal{A}$ of $X$ contains a finite subcollection that also covers $X$, then $X$ is called a $\textbf{compact metric space}$.
(For $\textbf{sequentially compactess}$)
Let $(X,d)$ be a metric space, let $E\subseteq X$, if every sequence of points of $E$ has a convergent subsequence, then $E$ is called a $\textbf{sequentially compact set}$.
Let $(X,d)$ be a metric space. If every sequence of points of $X$ has a convergent subsequence, then $X$ is called a $\textbf{sequentially compact metric space}$.
I don't know which is the regular definition people use. And also, I'm not sure whether this two versions of definition differs in some tricky circumstance, just like the "open set" mentioned above.