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In Dugundji's book, chapter $V$, problems on section $1$, the $8$th problem says:

$Y$ is connected if and only if every open covering $\{U_{\alpha}\,|\,\alpha\in A\}$ of $Y$ has the following property: For each pair of sets $U_{\alpha_1},U_{\alpha_n}$, there are finitely many $U_{\alpha_2},\dots,U_{\alpha_{n-1}}$ such that $U_{\alpha_i}\cap U_{\alpha_{i+1}}\not=\emptyset$, $i=1,\dots,n-1$.

What I've paraphrased from this, is that if your space is connected then you can make a bridge from open set to open set using a finite number of open sets, something like this:

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Proving the necessity wasn't that dificult, but proving the sufficiency is not very easy, mostly because I can't get to find a way to prove that you can find just $n$ open sets. If $Y$ is connected it's not necessarely compact, so I can't use that, and I can't figure a way to find a contradiction if we assume there is a covering which doesn't have the property, a lot of things can happen in that case, one of them is that there could be a infinite number of $U_{\alpha}$'s. Wouldn't we need more hypothesis?

Ana Galois
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HINT: Assume that $Y$ is connected, and let $\mathscr{U}$ be an open cover of $Y$. Define a relation $\sim$ on $\mathscr{U}$ as follows: for $V,W\in\mathscr{U}$, $V\sim W$ if and only if there are $U_0,\ldots,U_n\in\mathscr{U}$ such that $U_0=V$, $U_n=W$, and $U_k\cap U_{k+1}\ne\varnothing$ for $k=0,\ldots,n-1$.

  • Show that $\sim$ is an equivalence relation on $\mathscr{U}$.
  • Let $\mathscr{C}$ be a $\sim$-equivalence class; show that $\bigcup\mathscr{C}$ is clopen in $Y$.
  • Conclude that there is only one $\sim$-equivalence class.
Brian M. Scott
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  • wow, this is just... elegant – Ana Galois Mar 19 '15 at 17:35
  • @ Brain : Your second statement says that $\mathscr C$ be a eqivalnce class, then Show that $\cup_{ G \in \mathscr C} G$ is clopen in $Y$ : – user120386 Oct 13 '15 at 08:31
  • Clearly $K$ is open in $Y$ and $x \in (Y - k)$, then there exist an element of the open cover which not belongs this class says $G$, let if possible $G \cap K \neq \phi$, then there $G' \in \mathscr C$ such that $G\cap G' \neq \phi$ , thus $G \in \mathscr C$ which is a contradiction. , So $Y-K$ is open then $K$ is closed .Am i Right – user120386 Oct 13 '15 at 08:43
  • @user120386: Yes, that’s right. – Brian M. Scott Oct 13 '15 at 16:47