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Let E be a collection of connected subsets $E_j$, all having a point in common. Prove that $\cup E_j$ is connected.

Here is my proof; please let me know if you agree:

Suppose E is disconnected and let $f$ be a continous map onto the discrete space {$0,1$}. Then $A=f^{-1}(0)$ and $B=f^{-1}(1)$ are disjoint, nonempty and $E=A\cup B$. Since each of the $E_j$ have a point in common, then either $\cup E_j \in A \text{ }or \cup E_j \in B$. Suppose, WLOG, that $\cup E_j \in A$. Then $B$ is empty, a contradiction. Hence, $\cup E_j$ is connected.

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    I take it you are trying to say, if $x$ is a common point of every $E_j$, without loss of generality we may assume $x \in A$, whence by connectedness of each $E_j$ we must have $E_j \subset A$ for every $j$. Therefore $\bigcup E_j \subset A$. – 2017-02-25

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I think you should expand on a few things in the proof. An immediate red flag that you aren't being very rigorous is when you don't explicitly mention the assumptions of the proof in your logic. You never referred to the fact that each $E_j$ is connected. Moreover, you referred to a function $f$ and then didn't really use it. That's fine, you don't really need to phrase this short proof in terms of continuous functions, but you should pick one approach and stick to it.

Starting from the beginning, I would also not start by assuming the existence of such a continuous function $f$. Why should it exist? This function exists because you are assuming $E$ is disconnected, and is therefore the disjoint union of nonempty open sets $A$ and $B$. 'Open' is important because then you can verify that the function $f$ mapping $A \rightarrow 0$ and $B \rightarrow 1$ is continuous (as the preimages of $\{0\}, \{1\}$, and $\{0,1\}$ ($A,B$, and $E$, respectively) are open by construction.

Now you can use this continuous function $f$ to be rigorous. Continuous functions map connected sets to connected sets, so $f(E_j) = 0$ or $f(E_j) = 1$. Therefore the image under $f$ of $E_j$ is determined by any point of $E_j$. Since we assume the $E_j$ have nonempty intersection, pick a point $x$ in their intersection to see $f(E_i) = x = f(E_j)$ for all $i,j$. Then $f(\bigcup_j E_j) = \bigcup_j f(E_j) = \bigcup_j f(x) = f(x)$.

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I think you are mixing two different proofs. One of which is answered by @Badam Baplan.

I will give you second proof.

Suppose $\cup E_j$ is disconnected. Then $\cup E_j=A \cup B$ where $A,B$ are open, non-empty and $A \cap B = \emptyset$.

If there exists $E_{j_3}$ such that $E_{j_3} \cap A \neq \emptyset$ and $E_{j_3} \cap B \neq \emptyset$, then $E_{j_3}=(E_{j_3} \cap A) \cup (E_{j_3} \cap B)$ is a separation of $E_{j_3}$ (Verify) which is a contradiction since $E_{j_3}$ is given connected in hypothesis.

Let $E_{j_1} \subset A$ and $E_{j_2} \subset B$. Then $E_{j_1} \cap E_{j_2} = \emptyset$. Again a contradiction to the fact that there is a point common to both $E_{j_1}$ and $E_{j_2}$.