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I want to show that $\mathbb R$ is not homeomorphic to $\mathbb R$ \{1}. An idea was to use the Intermediate Value Theorem, but I cannot seem to apply it.I would appreciate the help.

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Hint. Do you remember what the topological definition of a "connected space" is? Is $\mathbb{R}$ connected? Is $\mathbb{R}\setminus\{1\}$ connected? Can a connected space be homeomorphic to a non-connected one?

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    Can it be shown without using the notion of connectedness? @MikeHaskel – 2017-02-26
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    @user123 You have the Intermediate Value Theorem; this is basically a statement about the connectedness of intervals in $\mathbb{R}$ and the fact that the image of connected set is connected. So if $f: \mathbb{R} \to \mathbb{R}\setminus \{1\}$ is a homeomorphism, then there are $x,y\in \mathbb{R}$ so that $f(x) < 1< f(y)$. But then IVT implies there is $z$ between $x$ and $y$ so that $f(z)=1$. – 2017-02-26
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    @Hayden Then how can you show the contradiction. That there does not exist $f(z)=1$. Or it can be deduced directly since 1 is not part of the image. – 2017-02-26
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    The image is contained in $\mathbb{R}\setminus \{1\}$, so can there be such a $z$? – 2017-02-26