Let $ X,Y $ are two Banach space and $ T:X\to Y $ is a linear operator. Assume that the kernel of $ T $, denoted by $ N(T)=\{x\in X:Tx=0\} $ is closed. I want to find a counterexample such that for this case, $ T $ is not bounded.
I know that such result is true if $ Y=\mathbb{C} $ but I do not know how to continue. Can you give me some hints or references?
On
While I don't disagree with Robert Israel's answer, there is an issue of terminology. In discussing 'unbounded operators' on Banach spaces, it is usual to mean operators that are defined on a dense subspace of the source space $X$. In that context, unbounded operators may be defined without the axiom of choice. If $T$ is any unbounded closed operator, the kernel of $T$ is closed.
Example: on $H=\ell^2(\mathbb{N}^+)$, with the standard basis $\{e_n\}_{n=1}^\infty$. Let $T:H\to H$ be defined via $T(e_n) = n e_n$. The maximal 'natural' domain of $T$ is $$V=\left\{\vec{a}\in H: \sum_{n=1}^\infty n^2 |a_n|^2<+\infty\right\}$$ $V$ is a dense subspace of $H$ (contains the linear span of $\{e_n\}_{n=1}^\infty$), and $T$ is an unbounded closed operator on $V$. The kernel of $T$ is closed, in fact $\{0\}$.
Of course, $T$ is not defined on all of $H$ and to extend to all of $H$ you need the axiom of choice. It was proved by Andreas Blass in 1984 (here) that the existence of a linear basis to every vector space over any field implies the axiom of choice - although if you read the proof, the fields used there are hardly anything like $\Bbb{R}$ or $\Bbb{C}$.
Let $X$ be any infinite-dimensional Banach space, and $Y=X$. Let $B$ be a Hamel basis of $X$, i.e. a basis of $X$ as a vector space, so that every member of $X$ can be represented uniquely as a (finite) linear combination of members of $B$. Existence of such a basis follows from the Axiom of Choice. Let $x_1, x_2, \ldots$ be a sequence of distinct elements of $B$. Define $T$ so that $T x_i = i x_i$, and $Tb = b$ for every other member of $B$. Extend by linearity to all of $X$. Note that $N(T) = \{0\}$.
You're not going to find a much more explicit counterexample.
By a theorem of Solovay, it's consistent with ZF (the Zermelo-Frankel axioms, without Choice) + DC (Dependent Choice) that every subset of a complete separable metric space has the property of Baire. From this it would follow that every everywhere-defined linear operator from $X$ to $Y$, where $X$ and $Y$ are Fréchet spaces and there exist enough linear functionals to separate points of $Y$, is continuous. See J.D.M. Wright, BAMS 79 (1973) 1247-1250, Theorem 5.
So to get an unbounded operator you need Axiom of Choice (or something pretty close to it).
[EDIT] If, as in John Doe's answer, you are interested in operators defined on a dense subspace of $X$ rather than all of $X$, Goldberg and Kruse showed that there is a one-to-one compact operator $K: Y \to X$ with dense range if $X$ and $Y$ are infinite-dimensional Banach spaces, $X$ is separable and $Y'$ has a denumerable total subset. If so, $K^{-1}$ (with domain the range of $K$) is an unbounded densely-defined operator from $X$ to $Y$.