As you said $\bar{A} = A \cup \operatorname{Bd}(A)$, but there's a more stronger statement that you can and you should prove: $\bar{A} = \operatorname{Int}(A) \cup \operatorname{Bd}(A)$.
Now it should be fairly easy to prove that $\operatorname{Bd}(A) = \emptyset \iff \text{A is closed and open}$:
(1)If $\operatorname{Bd}(A)=\emptyset$
Then $\bar{A}=\operatorname{Int}(A)$ and since $\operatorname{Int}(A)\subset A \subset \bar{A}$ we conclude that $\operatorname{Int}(A) = A = \bar{A}$.
$\operatorname{Int}(A) = A$ shows that $A$ is open and $A = \bar{A}$ shows that $A$ is closed.
(2)If $A$ is closed and open.
A is open, then $A=\operatorname{Int}(A)$ and since $\bar{A} = \operatorname{Int}(A) \cup \operatorname{Bd}(A)$ we see that $\bar{A} = A \cup \operatorname{Bd}(A)$
But $A$ is also closed!, this means that $\bar{A} = A$ therefore $A = A \cup \operatorname{Bd}(A)$.
Now remember that $\operatorname{Int}(A) \cap \operatorname{Bd}(A) = \emptyset$, then $A \cap \operatorname{Bd}(A) = \emptyset$.
Because of $A = A \cup \operatorname{Bd}(A)$ and $A \cap \operatorname{Bd}(A) = \emptyset$ we conclude $\operatorname{Bd}(A) = \emptyset$
The converse does not hold. If conditions 1 and 2 hold, then $C$ is the connected component of $p$, but it is not necessarily the case that every point outside the component can be separated from $p$ by open sets.
To see that 1 and 2 imply that $C$ is the connected component of $p$, let $K$ be the connected component of $p$ in $X$. We need to show that $K = C$. Since by definition connected components are maximal connected sets, and $C \cup K$ is a connected set (the union of two connected sets with nonempty intersection is connected) containing $K$, it follows that $C \cup K = K$, i.e. $C \subset K$. For the revers inclusion, consider $q \in X \setminus C$. By assumption 2, there are disjoint open $U,V$ such that $p \in U$, $q\in V$, and $X = U \cup V$. Then $(U \cap K)$ and $(V \cap K)$ are disjoint relatively open (in $K$) sets with $K = (U \cap K) \cup (V \cap K)$, and by the connectedness of $K$ it follows that $U \cap K = \varnothing$ or $V \cap K = \varnothing$. Since $p \in U \cap K$ we must have $V \cap K = \varnothing$, in particular $q \notin K$. Since this holds for all $q \in X \setminus C$, the inclusion $K \subset C$, and consequently $K = C$, is proved.
To see that the converse does not hold, an example suffices. Let
$$X = \{(0,0),(0,1)\} \cup \bigcup_{n = 1}^{\infty} \bigl\{ (1/n,t) : 0 < t < 1\bigr\}$$
with the subspace topology inherited from $\mathbb{R}^2$. Then the connected component of $p = (0,0)$ in $X$ is the singleton $\{(0,0)\}$, but $q = (0,1)$ cannot be separated from $p$ by open sets. The quasicomponent of $p$ is the two-element set $\{p,q\}$.
By definition, the quasicomponent of a point is the intersection of all clopen (that is, closed and open) sets containing that point. The quasicomponent of $p$ consists of exactly those points of $X$ that cannot be separated from $p$ by open sets. For if $r$ does not belong to the quasicomponent of $p$, then there is a clopen set $A$ containing $p$ but not $r$, then $U = A$ and $V = X\setminus A$ is a decomposition of $X$ into two disjoint open sets that separates $p$ from $r$. And if $r$ can be separated from $p$ by open sets $U,V$, then these two sets are actually clopen, so $r$ does not belong to the quasicomponent of $p$.
To see that in the above example $q$ belongs to the quasicomponent of $p$, consider a clopen $A$ containing $p$. Since $A$ is open, there is an $N$ such that $A \cap \{(1/n,t) : 0 < t < 1\} \neq \varnothing$ for all $n \geqslant N$. Since the segment $L_n = \{(1/n,t) : 0 < t < 1\}$ is connected (even path-connected), and $L_n \cap A,\, L_n \setminus A$ is a decomposition of $L_n$ into disjoint open sets, one of these must be empty, thus $L_n \subset A$ for all $n \geqslant N$. But $q$ is the limit of the sequence $\bigl((1/n, 1-1/n)\bigr)_{n \geqslant N + 1}$ all of whose points belong to $A$ by what we've seen, and $A$ is closed, so it follows that $q \in A$. Showing that no other point belongs to the quasicomponent of $p$ is easy: $X \setminus L_n$ is a clopen set containing $p$ but none of the points on $L_n$.
And finally, since $\{p,q\}$ is not connected, it follows that the component of $p$ is a proper subset of its quasicomponent.
Best Answer
This is not true generally unless $\overline{A\cap B}=\overline{A}\cap \overline{B}$. \begin{align} \partial (A\cap B)&= \overline{A\cap B}-(A\cap B)^{o} \\ &=(\overline{A}\cap \overline{B})-(A^{o}\cap B^{o}) \\ &=(\overline{A}\cap \overline{B})\cap(A^{o}\cap B^{o})^c \\ &=(\overline{A}\cap \overline{B})\cap(A^{o^c}\cup B^{o^c}) \\ &=(\overline{A}\cap \overline{B}\cap A^{o^c})\cup(\overline{A}\cap \overline{B}\cap B^{o^c}) \\ &=(\overline{A}\cap \partial B)\cup (\overline{B}\cap \partial A) \end{align} By this post, $\overline{A\cap B}=\overline{A}\cap \overline{B}$ implies discrete space. So this is impossible in $\Bbb{R}^n$. However, we can prove generally $$ \partial(A\cap B)\subset (\bar A \cap \partial B) \cup (\partial A \cap \bar B) $$ for it is always true that $\overline{A\cap B}\subset \overline{A}\cap \overline{B}$. This can be done easily by replacing "$=$" with "$\subset $" at 2nd line of above proof and rest follows.