Give an example to show that converse is not true: I know that  if  $ T: X \rightarrow Y$ be a continuous then $Gr(T)=\{(x,Tx):x\in X\}$ is closed in $X\times Y$.
Since let $({x_n,T(x_n)}) \subseteq Gr(T)$ such that $(x_n,T(x_n)) \rightarrow (x,y) \in $$X\times Y$.Thus $x_n\rightarrow x$ , $T(x_n) \rightarrow y$.
But $T$ is continuous.So $T(x_n)\rightarrow Tx$.
i.e  $y=T(x)$ and $(x,y)\in Gr(T)$.
But that converse is not true.For example $ f(x) =
\begin{cases}
Sin\frac{1}{x} & x\neq0 \\
0   & x=0
\end{cases}$
is not Continuous but i can not prove that $Gr(f)$ is closed.
Can you show this...

It seems that you're not necessarily looking for a linear map.
Consider the example $f:[-1,1] \to \Bbb R$ given by
$$
f(x) = 
\begin{cases}
1/x & x \neq 0\\
0 & x = 0
\end{cases}
$$
Note that the graph of the function is closed, but the function is discontinuous.
Interesting fact: functions on a compact metric space are continuous if and only if their graph is compact.
