Understanding Fixed Points on $[0,1)$ and $(0,1)$ Munkres asks the following question:

Let $f:[0,1] \rightarrow [0,1]$ be continuous. Show that there exists a point $f(x) = x$ (called a fixed point). What if the domain and range are $[0,1)$ or $(0,1)$?

The arguement for the first part is to produce a function $g = f-x$, argue that $g$ is continuous and use intermediate value theorem. A counterexample for the second part is $f(x) = 1/2 + x/2$.
My question: I'm not understanding what fails in the second part of the question. Intermediate value theorem still holds because $[0,1)$ and $(0,1)$ are still members of the order topology - so I should be able to produce a fixed point. So why does a counterexample exist?
EDIT: The Theorem Reads as Follows:

Let $f:X \rightarrow Y$ be a continuous map, where $X$ is a connected space and $Y$ is an ordered set in the order topology. If $a$ and $b$ are two points in $X$ and $r$ is a point in $Y$ lying in between $f(a)$ and $f(b)$ then there exists a point $c \in X$ such that $f(c) = r$.

 A: Tsemo's answer is correct, but intuitively, it follows because when your interval is missing an endpoint, you can 'hide' the fixed point in the 'hole' formed by the missing endpoint.  For example, $f(x) = x^2$, viewed as a map from $(0,1)$ to $(0,1)$ is fixed point free, though of course, this is just because it's fixed points are 0 and 1, which are excluded from your domain.
A: The intermediate value is true for closed intervals, if you define $f(x)=1/2+x$ on $(0,1)$ and $g(x)=f(x)-x=1/2-x/2$, remark that $g(x)>0$, if you extend the domain of $f$ to $[0,1]$ $g(1)=0$, so it may happens that the element such that $f(x)=x$ where $f:[a,b]\rightarrow [a,b]$ is $a$ or $b$.
Read the proof of the intermediate value theorem, you will understand why the domain need to be closed. One way to prove that there exists $x\in [a,b]$ such that $f(x)=0$ if $f(a)f(b)<0$ is to construct a Cauchy sequence $x_n\in [a,b]$ such that $\lim\limits_nf(x_n)=0$, if $[a,b]$ is closed, $x=\lim\limits_nx_n\in[a,b]$ and since $f$ is continuous, $f(x)=\lim\limits_nf(x_n)=0$.
A: The different behavior of $[0, 1]$ and $[0, 1)$ lies in the initial information that you plug into the intermediate value theorem. You define 
$$
g(x):=f(x)-x, \qquad x\in I, 
$$
where $I$ is either $[0, 1]$ or $[0, 1)$. In the first case, the extremes $0$ and $1$ are part of the domain, so it makes sense to evaluate
$$
g(0)=f(0)\ge 0, \qquad g(1)=f(1)-1\le 0, $$
and this sign change gives you the existence of a zero of $g$, via the intermediate value theorem.
If, on the other hand, $I=[0, 1)$, then it does not make sense to evaluate $g$ at $1$. You can only evaluate it at points $1-\epsilon$, with $\epsilon >0$ as small as you wish but not zero. 
In the counterexample you gave, that is, $f(x)=\tfrac{x}2-\tfrac12$, it turns out that $g(1-\epsilon)> 0$ for all $\epsilon>0$. There is no way to induce the sign change necessary to prime the intermediate value theorem.
