Show that if the intersection is non-empty, then the union is connected of the following [duplicate]

This question already has an answer here:

Let $\{A_i\}_{i \in I}$ be a family of connected subsets of a metric space $X$ ($I$ is some set of indices). Show that if the intersection $\bigcap A_i \neq \emptyset$ , then $\bigcup A_i$ is connected.

marked as duplicate by user147263, Ilmari Karonen, Jack Lee, Brian M. Scott general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 4 '15 at 1:22

Since $\bigcap A_i \ne \emptyset$, let $x \in \bigcap A_i$. Now if $U$ and $V$ are non-empty disjoint open sets separating $\bigcup A_i$ then $x \in U$ or $x \in V$. Without loss of generality, suppose $x \in U$. Now, where does $x$ come from and what can we say about it?
• Meaning $A = U\cup V$. – Kevin Sheng Nov 4 '15 at 0:14