Recall that an operator is closed if a sequence $(x_{n})\subset D(T)$ converges to some $x\in X$ and $Tx_{n}$ converges to $y\in Y$ implies $x\in D(T)$. Equivalently, the graph of $T$ is closed in the direct sum $X\oplus Y$.
Proof of i). Since $T$ is injective, we can define a linear operator $T^{-1}$ on the range of $T$, denoted $R(T)$, by $T^{-1}y=x$, where $y=Tx$. Let $y_{n}=(Tx_{n})\subset D(T^{-1})=R(T)$ be a sequence such that $y_{n}\rightarrow y\in Y$ and $T^{-1}y_{n}\rightarrow x\in X$. But since $y_{n}=Tx_{n}$ for a sequence $(x_{n})\subset D(T)$, it follows from the hypothesis that $T$ is closed that $x\in D(T)$ and $y=Tx$. Whence, $x=T^{-1}y$.
Proof of ii). To see that $T$ is not closed, consider the sequence formed by the truncations to the first $n$ components of a suitable element $x\in\ell^{p}$.
Define an operator $\bar{T}:D(\bar{T})\rightarrow\ell^{p}$ as follows. Let
$$D(\bar{T}):=\left\{x=(x_{k})\in\ell^{p} : (kx_{k})_{k}\in\ell^{p}\right\}$$
and define $\bar{T}x=(kx_{k})_{k}$. It is evident that $\bar{T}$ is linear and is an extension of $T$.
To see that $\bar{T}$ is closed, let $x^{(n)}=(x_{k}^{(n)})_{k}$ be a sequence in $D(\bar{T})$ converging to $x\in\ell^{p}$ such that $Tx^{n}\rightarrow y\in\ell^{p}$. I claim that $x=(k^{-1}y_{k})_{k}$. Indeed,
$$\sum_{k}\left|x_{k}^{(n)}-k^{-1}y_{k}\right|^{p}\leq\sum_{k}k^{p}\left|x_{k}^{(n)}-k^{-1}y_{k}\right|^{p}=\sum_{k}\left|kx_{k}^{(n)}-y_{k}\right|^{p}\rightarrow 0, \ n\rightarrow\infty$$
The claim follows from the uniqueness of limits. Clearly, $x\in\ell^{p}$ and $(kx_{k})\in\ell^{p}$, whence $x\in D(\bar{T})$ and $\bar{T}x=y$.
