I had some new thoughts on this that are too long for a comment. I have a proof sketch if the space is path-connected, but it would only work if you can show that the space can't contain a Y-shaped graph.
I will use the fact that any injective map of the interval into a metric space is an embedding.
Assuming the space is path-connected, choose $x,y$ in the space, and let $\alpha$ be an arc connecting them. Let $\beta$ be a path (not necessarily an arc) from $x$ to $y$ that only intersects each endpoint once and that does not lie entirely in the image of $\alpha$ (if such a path exists). I claim that $\beta$ and $\alpha$ intersect only in their end points $x$ and $y$. Otherwise, there exists a point $p$ in the images of both $\alpha$ and $\beta$ such that there is an $\epsilon>0$ with $\beta((\beta^{-1}(p),\beta^{-1}(p)+\epsilon)$ is disjoint from $\alpha$. You choose $p$ to be the first point after $x$ where $\beta$ leaves $\alpha$.
Let $\gamma=\beta([\beta^{-1}(p),\beta^{-1}(p)+\frac{1}{2}\epsilon]$. Then $\gamma\cup\alpha$ is an embedded space with a branch point, which is not possible as discussed above.
Thus, any two paths from $x$ to $y$, each intersecting the endpoints only once, must be disjoint except in their endpoints. We can choose $\beta$ to be an arc by shrinking it; it will still be disjoint from $A$. There cannot be three such arcs, since we would get another branch point. Thus, if there are two points where two such arcs exist, the arcs must be surjective and we have a circle.
If there is only one path between each pair of points, then the space is the union of closed intervals pasted along closed intervals, and is an interval.
Edit: This is just a proof sketch; some of the details may need hammering out.
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
Fix $x_0 \in X $. Then, the continuous(!) map $$ \Phi: X \to \Bbb {R}, x \mapsto d (x,x_0) $$ has an (at most) countable, connected image.
Thus, the image is a whole (nontrivial!, if $X $ has more than one point) interval, in contradiction to being countable.
EDIT: On a related note, this even show's that every connected metric space with more than one point has at least the cardinality of the continuum.