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 $$X\times Y$.Thus $x_n\rightarrow x$ , $T(x_n) \rightarrow y$.
But $T$ is continuous.So $T(x_n)\rightarrow Tx$.
i.e  $y=T(x)$ and $(x,y)\in Gr(T)$.
But that converse is not true.For example $ f(x) =
\begin{cases}
Sin\frac{1}{x} & x\neq0 \\
0   & x=0
\end{cases}$
is not Continuous but i can not prove that $Gr(f)$ is closed.
Can you show this...
 A: It seems that you're not necessarily looking for a linear map.
Consider the example $f:[-1,1] \to \Bbb R$ given by
$$
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.
A: Hint. Let $x_n := \bigl(\frac \pi 2 + 2n\pi\bigr)^{-1}$. And $y_n := f(x_n)$. Does $x_n$ converge? Does $y_n$ converge? If so, and $x$ and $y$ are the limits, do we have $y=f(x)$?
A: Quite often, closed graph is equivalent to continuity:
For linear maps between Banach (or Frechet) spaces this is Banach's classical closed graph theorem.
For maps $f:X\to Y$ from a metric space $X$ to a compact metric space $Y$, this is quite elementary: If $x_n\to x_0$ one checks $y_n=f(x_n)\to f(x_0)$ by using every subsequence $y_n'$ has a further subsequence $y_n''\to f(x_0)$: By compactness of $Y$ there is a subsequence $y_n''\to y_0$ for some $y_0\in Y$. But, if $x_n''$ denotes the corresponding subsequence of $x_n$, one has $(x_n'',y_n'')\in Gr(f)$ which implies $(x_0,y_0)\in \overline{Gr(f)}= Gr(f)$ and hence $y_0=f(x_0)$. 
