TITLE: Proof about a Topological space being arc connect
QUESTION [5 upvotes]: While reading a book i found a topological space described as:

Let $(X,\tau)$ be the topological space formed by adding to the
  ordinary closed unit interval $[0,1]$ another right end point,say
  $1*$, with the sets $(a,1)\cup${1*} as a local neighborhood basis.

Then it says that such topological space is arc connected.
I found almost exactly the same question here which has yet to be solve,however i'll provide some details.

The book itself states that since [0,1] and [0,1)$\cup${1} are homeomorphic as subspaces,and the subspace topology on [0,1] is
  Euclidean,X is the union of two compact subspaces and thus compact,by
  the same reasoning it is arc connected.*

How can such argument prove me that there is a injective path from $1$ to $1*$? Is it possible to explicit such path?
Further details: As the book states:

Path and arc connectednesss relate to the existence of certain
  continuous functions from the unit interval into a topological
  space.Continuous functions from the unit interval are called paths,if
  they are one-to-one they are arcs.

REPLY [4 votes]: This space is not arc-connected.  Indeed, suppose $f:[0,1]\to X$ is an arc such that $f(0)=1$ and $f(1)=1^*$.  Then $f(1/2)\in [0,1)$, and $f|_{[0,1/2]}$ is a (reparametrized) path from $1$ to $f(1/2)$ in $[0,1]$.  Thus $f([0,1/2])$ must contain all of $[f(1/2),1)$.  But by a similar argument, $f([1/2,1])$ also must contain all of $[f(1/2),1)$.  This contradicts injectivity of $f$.