The following problem can be obtained in the book 'An Introductory Course in Functional Analysis' by Bowers and Kalton, page $59$, problem $3.5$
Known facts about $j$: It is a linear, injection and isometry map.
Question: Let $X$ be a Banach space with bidual $X^{**}$ and let $j:X \rightarrow X^{**}$ be the natural embedding. Show that $j(X)$ is a closed subset of $X^{**}.$
My attempt: I wish to prove that $j(X)$ is sequantially closed.
Suppose that a sequence $(x^{**}_n)_{n \in \mathbb{N}}$ in $j(X)$ converges to $x^{**}$.
Since $j$ is injective, there exists a unique $x_n \in X$ such that $j(x_n) = x_n^{**}$ for all $n \in \mathbb{N}.$
I am having trouble to show that the sequence $(x_n)_{n \in \mathbb{N}}$ converges in $X.$ If it does converge, say converge to $x$, then by continuity of $j,$ we have $(x_n^{**})_{n \in \mathbb{N}}$ converges to $j(x)$, which is equal to $x^{**}.$
By the way, authors suggest that since $j$ is isometry, therefore we have $j(X)$ is closed in $X^{**}.$ But I don't see it.
UPDATE: The following is my full proof to the question. Is it correct?
We wish to show that $j(X)$ is sequentially closed in $X^{**}.$ Let $(x_n^{**})_{n \in \mathbb{N}}$ be a sequence in $j(X)$ that converges to $x^{**}.$ We wish to show that $x^{**} \in j(X).$ By injectivity of $j,$ for all $n \in \mathbb{N},$ there exists a unique element $x_n \in X$ such that $j(x_n) = x_n^{**}.$ Recall that any convergent sequence is Cauchy. In particular, $(x_n^{**})_{n \in \mathbb{N}}$ is a Cauchy sequence in $j(X).$ Since $j$ is an isometry, we have $(x_n)_{n \in \mathbb{N}}$ is Cauchy in $X.$ Since $X$ is a Banach space, $(x_n)_{n \in \mathbb{N}}$ converges to $x \in X.$ As $j$ is continuous, we have $(x_n^{**})_{n \in \mathbb{N}}$ converges to $j(x).$ By uniqueness of limit, we have $j(x) = x^{**}.$ Hence, $x^{**} \in j(X).$
$J(X)$ is an isometric embbeding of $X$ in $X^{**}$ thus because of the fact that $X$ is complete then $J(X)$ is also a complete subset of $X^{**}$ and also $X^{**}$ is a banach space thus $J(X)$ is closed.
In general ,remember that a complete subset of a complete space is closed.