Prove that $f$ has a fixed point . For $f:[a,b]\rightarrow [a,b]$ is a continuous . Prove that $f$ has a fixed point . Is that true if we change $[a,b]$ by $[a,b)$ or $(a,b)$.
 A: Another idea without using Intermediate Value Theorem:
Proof: Firstly, let us pick any point from $[a,b]$, say $x_0$. Assume we have defined $x_n$, let $x_{n+1}=f(x_n)$. Thus we abtain a sequence $\{x_n: n\in \Bbb N^+\}$ from $[a,b]$. Since $[a,b]$ is compact, then for any infinite subset of $[a,b]$ there exists an limit point in $[a,b]$, say $x$. 
Then we can claim that $f(x)=x$. Since $f$ is continuous, $f(x)=f(\lim x_n)=\lim f(x_n)=\lim x_{n+1}=x$. This complete the proof.

The interval $[a,b)$ or $(a,b)$ is not enough, Since $x$ maybe fell on the point $b$, which is no longer in the origin interval.

Added: The answer seems right. However it is not right, just as  Landscape points; he gives us a counterexample in the comments. Many times if the answer is wrong, I will delete it immediately. However this is a special case. From it, I learn some lessons. Something cannot be of course. So I hope to keep it as long as I can.
A: Hint: Apply Intermediate Value Theorem on the function $g(x) = f(x) - x$.
The closed interval is needed, because of the conditions of IVT. For example, the function $f:(0,\infty) \rightarrow ( 0, \infty)$ given by $f(x) = x + 1$ has no fixed point. You can modify this slightly if you want an interval $[a,b)$ or $(a,b)$.
A: I have an ideal .
1) We have $f:[0,1]\rightarrow [0,1]$ ($f$ is continuous mapping ) has a fixed point . ( from $x-f(x)=g(x)$ with $g(z)=0\Rightarrow z=f(z)$.
2) We need prove that : $X,Y$ be homeomorphic topological spaces. then each continous mapping $h:X\rightarrow X$ has a fixed point if and only if $k:Y\rightarrow Y$ has a fix point .
$Proff.$ 
We have $h:X\rightarrow X$ , $g:Y\rightarrow X$,$f:X\rightarrow Y$ , $k:Y\rightarrow Y$.Suppose $h$ has a fixed point , then $h=g.k.f$ is continuous $z\in X : h(z)=z$. Let $w=f(z)$. We have $k(w)=k(f(z))=fg(k(f(z)))=f(h(z))=f(z)=w.$. Thus $w$ is fixed point of $k$.
