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.
Some comments.
You pick up a connected component $C$, an element $x \in C$ and then provide a definition (by extension) of $C$ which is fine.
But then you use the theorem saying that a union of connected sets all containing the same point is connected. This is not necessary.
Better to come back to the definition of a connected set. Suppose that $C$ is the union of two (relative) disjoint open subsets $C_1,C_2$ with $x \in C_1$. Then for $x \neq y \in C$, it exists a connected set $C_y$ containing $y$ and included in $C$. Necessarily $y \in C_1$ as if not $C_y$ would be the disjoint union of the non empty (relative to $C_y$) open subsets $C_1 \cap C_y, C_2 \cap C_y$. Therefore $C_2 = \emptyset$, proving that $C$ is connected.
Best Answer
Use this fact: the closure of a connected set is connected. Here is the proof: if $\overline A=C\cup D$ for some non-empty sets $C,D$ open in $\overline A$. Since $A$ is connected, we may assume without loss of generality, that $A\subseteq C$. Then, $\overline A\subseteq \overline C.$ But $\overline C\cap D=\emptyset$ (why?) and so $\overline A\cap D=\emptyset$. Thus, we have shown that $D$ is empty, which is a contradiction.
Now, let $C$ be a component of $x$. Then, $x\in \overline C$, which is connected, so by definition of component, $\overline C\subseteq C$. That is, $\overline C=C$, so $C$ is closed.