general topology - 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 $f(c) = r$.

Answer

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.