Proof for closedness and surjection for the isometry on closed unit ball Let $D$ be the closed unit disc in $\Bbb R^2$. Let $d(.,.)$ denote the Euclidean distance in $D$. Let $T: D \to D$ be a mapping such that $d(T(x), T(y))=d(x,y)$ for all points $x$ and $y$ in $ D$. 


*

*There exists $x_0 \in D$ such that $T(x_0) = x_0$.

*The image of $T$ is closed.

*The image mapping $T$ is surjective.


How can I prove or disprove above three statements? Where I can start to think?
Any help is appreciated. Thank you.
 A: For $(2)$, we will prove a stronger statement, i.e. the image of an isometry $T$ of a compact metric space $X$ is closed. It's easy to verify that $T$ is continuous. As continuous maps preserve compactness, $T(X)$ is compact, hence closed (metric spaces are Hausdorff).   
For $(3)$, again, we will generalize the result to compact metric spaces, i.e. every isometry defined on a compact metric space is surjective. For the sake of contradiction, let's assume $a\notin T(X)$. By $(2)$, $T(X)$ is closed in $X$. Then there exists an open ball $B(a, \epsilon)$ of $a$ such that $B(a,\epsilon) \cap T(X)=\emptyset$. We are going to construct a sequence $\{x_n\}$ that doesn't have a convergent subsequence. Let $x_1=a, x_n=T(x_{n-1})$. Then it's easy to prove by induction that $d(x_n,x_m)\geq\epsilon$ for any $n$ and $m$. But a convergence subsequence must be Cauchy. Hence $X$ is not sequentially compact, hence not compact (compact$\Rightarrow$sequencially compact for matric spaces). Contradictory to our assumption.
For $(1)$, since $T$ is continuous, this follows from Brower fixed-point theorem.
