Question: is a closed disc is a example of closed surface.
I know that, the boundary of an open disk viewed as a manifold is empty, while its boundary in the sense of topological space is the circle surrounding the disk. as by definition of closed surface , A closed surface is a surface that is compact and without boundary. So i thought closed disc is not a closed surface, is am i correct.
Yes, it is not closed because it has a non-empty boundary. The open disc is not closed because it is not compact, even though it has empty boundary.
It may help in the latter case to recall the definition of a boundary of a surface as it appears you may be confused. For a surface $S$, a point $x\in X$ is a boundary point if there exists an open neighbourhood $U\subset S$ containing $x$ such that $U$ is homeomorphic to $\{(s,t)\in\mathbb{R}^2\mid t\geq 0\}$, the closed upper half plane. The boundary $\partial S$ of a surface $S$ is the set of all boundary points of $S$, so $\partial S=\{x\in S\mid x\mbox{ is a boundary point}\}$.
Unfortunately the word "closed" is used with two different meanings here.
When people started to think about surfaces as $2$-dimensional manifolds the word "compact" was not yet around. So they called compact $2$-dimensional manifolds (the $2$-sphere, $2$-torus, etc.) closed surfaces.
Nowadays the word "closed"in the first place refers to a property of subsets $A\subset X$ of a toplogical space $X$. Some subsets have this property, others don't. Given an arbitrary subset $A\subset X$ one can close it by forming its closure $\bar A=A\cup\partial A\supset A$. In this sense the closed unit disk is the closure $\bar D$ of $D:=\{(x,y)\in{\mathbb R}^2\>|\>x^2+y^2<1\}$.
