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I think this should be true. If it's indeed the case, it seems like this should be a known result, so references are welcome.

I managed to prove that (assuming $S$ is the connected subset) $S$ minus one of the three small triangles (but don't remove the two points which connect the small triangle to the rest) is also connected. This can be stated more generally for connected subsets in three spaces cyclically connected by three points. (basically, if $S$ restricted to one of them fails to be connected, then the restriction to the other ones has to be connected)

I'm not sure how to continue. Additionally, $S$ doesn't have to be closed, which causes further problems. Any ideas are also welcome.

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    $\begingroup$ It's not even obvious to me that the triangle itself is path connected. Do you have a reference or proof for this? I don't see what to do about the points that don't lie on the boundary of any removed triangle. $\endgroup$
    – MartianInvader
    Oct 21, 2014 at 22:21
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    $\begingroup$ @MartianInvader Path connectedness of the whole triangle is quite easy to see directly - you just start converging to the point if it's not on some trivial line. Anyway, it's a locally connected continuum, i.e. a Peano continuum, which is in fact a continuous image of the interval [0,1]. $\endgroup$
    – user2345215
    Oct 21, 2014 at 22:27

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Perhaps surprisingly, this is false. And it's quite an old result of Knaster and Kuratowski: http://www.ams.org/journals/bull/1927-33-01/S0002-9904-1927-04326-9/S0002-9904-1927-04326-9.pdf

In short, let $V$ be the set of vertices of all triangles appearing in the standard construction of the Sierpiński triangle. And let $B$ be any Bernstein subset of the Sierpiński triangle, i.e. a set intersecting, but not containing, every perfect subset of the triangle. Transfinite recursion provides an easy construction since every perfect subset has the cardinality of the whole triangle.

It's easy to see that $B\cup V$ contains no arc. It turns out it's connected, which is the hard part.

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    $\begingroup$ At least that explains why kept running into problems :) $\endgroup$
    – Hagen von Eitzen
    Feb 19, 2016 at 17:04

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