If $x \in \operatorname{cl}(A)$, where $A$ is a connected subspace of a topological space $X$, then $A \cup \{x\}$ is connected. 
Prove that if $x \in \operatorname{cl}_X(A)$, where $A$ is a connected
  subspace of a topological space $X$, then $A \cup \{x\}$ is connected.

My attempt:
Suppose, in order to find a contradiction, that $B \cup C = A  \cup \{x\}$ where $B,C$ are open in $A \cup \{x\}$ non empty and disjoint. 
Then, we have either $A \subseteq B$ or $A \subseteq C$. Indeed, if this wouldn't be the case, then both $A \cap C$ and $A \cap B$ are non empty, and then 
$A = A \cap (B \cup C)  = (A \cap B) \cup (A \cap C)$ gives a union of disjoint non empty open (in A) sets, contradicting the connectedness of $A$.
Without loss of generality, we may assume that $A \subseteq B$. Then $x \in C$, or else $A \cup \{x\} = B$, meaning that $C = \emptyset$, which isn't possible.
Write $C = G \cap (A\cup \{x\})$ with $G$ open in $X$. Because $x \in \operatorname{cl}(A)$, and $x \in G$, it follows that $A \cap G \neq \emptyset$. Pick $y \in A \cap G$. Then $y \in A \subseteq B$ and $y \in A \cap G \subseteq C$, so $y \in B \cap C$. This is the desired contradiction.
Is this correct? 


The OP's proof is correct, but by changing the last 2 paragraphs it
  can be nailed down in another way.

Suppose, in order to find a contradiction, that $B \cup C = A  \cup \{x\}$ where $B,C$ are open in $A \cup \{x\}$ non empty and disjoint. 
Then, we have either $A \subseteq B$ or $A \subseteq C$. Indeed, if this wouldn't be the case, then both $A \cap C$ and $A \cap B$ are non empty, and then 
$A = A \cap (B \cup C)  = (A \cap B) \cup (A \cap C)$ gives a union of disjoint non empty open (in A) sets, contradicting the connectedness of $A$.
Without loss of generality, we may assume that $A \subseteq B$. But then it must be that $B = A$ and $C$ is the open singleton $\{x\}$.
Since $x \in \operatorname{cl}(A)$, any open set in $A \cup \{x\}$ that contains $x$ can't be a singleton. This is the desired contradiction.
