I think you have the right idea! In your proof you say, "assume that there exists a $B_1$ such that $D\cap B_1=\varnothing.$ Then $D \cap B = \varnothing$. Is this the same $B_1$ as before? How do you draw this conclusion?
Perhaps, if you want to draw a contradiction, assume that $B_1, B_2 \subset B$ are open, disjoint, and non-empty such that $B_1 \cup B_2 = B$. Now, since $C$ is connected, either $B_1 \cap C = \varnothing$ or $B_2 \cap C = \varnothing$ (why!?). Without losing generality, suppose that $B_1 \cap C = \varnothing$. Then $C \subset B_1^c$. However, $B_1^c$ is a closed set containing $C$ and therefore $\overline{C} \subset B_1^c$ meaning that $B_1 \not\subset \overline{C}$. However, this creates a contradiction (why!?).
Your vague proof idea using suprema (Least upper bounds) is the basis for the proof of connectedness of ordered spaces:
Note that $\mathbb{R}$ is an ordered space in the sense that the topology is generated by all open sets of the form $L_a = \{x \in X: x < a\}$ and $R_a = \{x \in X: x > a\}$ , where $a \in X$. (Note that this means that the topology also contains all intervals $(a,b) = L_b \cap R_a$ as well, and the intervals with the $L_a$ and $R_a$ form a base for $X$).
An ordered topological space is connected iff it has no gaps and no jumps, which can also be stated as that it is densely ordered (for every $x < y$ in $X$ we have $z \in X$ with $x < z < y$; that's the no jumps part) and order complete (every set $A$ that is bounded above in $X$ has a least upper bound $\sup A \in X$; this is the no gaps property).
It is well-known that $\mathbb{R}$ has both of these properties, while $\mathbb{Q}$ fails the second and $\mathbb{Z}$ fails the first (and so both are disconnected ordered topological spaces).
Necessity:
Suppose $X$ has a gap $x <y$ with no points strictly in-between, then $L_y$ and $R_x$ are disjoint and non-empty, open (definition of order topology) and cover $X$ so $X$ is disconnected.
If $A$ is a set with upperbound $a_0$ but no supremum in $X$, then define $U = \{x \in X: \exists a \in A: x < a\}$, and $V = \{x \in X: \forall a \in A: a \le x\}$. $V$ is the set of upperbounds of $A$ (and this set has the property $v \in V, x > b$ then $x \in V$) and $U$ is the set of non-upperbounds of $A$ (and if $u \in U, x < u$ then $x \in U$). So by definition $U \cup V = X$.
$U$ is open, because if $u \in U$, $u < a$ for some $a \in A$ and then $u \in L_a \subseteq U$ and so $u$ is an interior point of $U$. $V$ is also open, because if $v \in B$ it's an upperbound of $A$ and there is no smallest one, so we have some smaller $b < v$ which is also an upperbound of $A$ and then $v \in R_{b} \subseteq V$, and $v$ is also an interior point of $V$. As $a_0 \in V$ and any $a \in A$ is in $U$ (or it would be a maximum, hence supemum of $A$), both sets are non-empty. So then $X$ is also disconnected.
Sufficiency
Suppose $X$ has no gaps and jumps. Assume for a contradiction that $X$ is disconnected, so $X = U \cup V$, where $U$ and $V$ are non-empty, open and disjoint. We can pick $u_0 \in U$ and $v_0 \in V$ such that $u_0 < v_0$ (we rename $U$ and $V$ if necessary). Define $U_0 = U \cap [u_0,v_0]$ and $V_0 = V \cap [u_0, v_0]$ (which are both open in $[u_0,v_0]$) and as $U_0$ is bounded above by $v_0$, $s = \sup U_0$ exists. Note that $s \in [u_0, v_0]$ ($s \le v_0$ is clear ($v_0$ is an upperbound of $U_0$, $s$ the smallest one) and $u_0 \le s$ ($s$ is an upperbound for all elements of $U_0$ so also of $u_0$). So $s \in U_0$ or $s \in V_0$.
Suppose $s \in V_0 (\subseteq V)$. Then there is an interval $(l,r)$ of $X$ such that $s \in (l,r) \subseteq V$, as $V$ is open. As $l < c$, and $l < s \le v_0$, $l$ cannot be an upperbound for $U_0$ so we have some $u \in U_0$ with $l < u$ (As $u \le s$ by definition, $u \in (l,r)$ so $u \in V$, contradiction. So $s \notin V_0$.
So then $s \in U_0$. As $U$ is open we have some interval $(l,r)$ again, such that $s \in (l,r) \subseteq U$ (clearly $r \le v_0$, or $v_0 \in U$). So $s < r$ and we find some $t$ with $s < t < r$ by the "no jumps" property. $t \in (l,r)$ so $t \in U$, but then $s$ is not even an upperbound for $U_0$ as $ t \in U_0$ and $ t > s$, contradiction. So the assumption that $X$ was disconnected was false. So a space $X$ with no gaps or jumps is connected.
Your "argument" above is much to vague ("I draw all the possibilities"..etc.; a picture (had you even included it) does not a proof make, but strict reasoning does, where pictures can assist the intuition).
Best Answer
$c$ cannot be the right limit of $I$ because $c<b$ and $b\in I$.
$c\notin U$ because $c$ is the supremum of $U$, which is open. In other words, if $c\in U$, then some neighborhood of $c$ would be included in $U$, but then there would have to be a value greater than $c$ in $U$, which would mean that $c$ wasn't the supremum.