Image of the canonical map is closed in a non-reflexive space Let $X$ be a Banach space. We have $\tau : X \rightarrow$ $X^{**}$ is the canonical map, isomorphic and isometric but suppose that X is not reflexive, so the map is not surjective. Now $\tau (X)$ is a closed subspace of $X^{**}$. Why can I say that $\tau(X)$ is closed? Is it something related to the map?

$\tau$ is an isometry so its range is complete. This implies that it is closed in $X^{**}$.

More simply argue as follows: Let $\tau(x_n) \to x^{**}$ in $X^{**}$. Then $\|x_n-x_m\| =\|\tau (x_n)-\tau (x_m)\|\to 0$. Since $X$ is complete there exists $x \in X$ such that $x_n \to x$. Since $\tau$ is continuous we get $\tau (x_n) \to \tau (x)$. Hence $x^{**}=\tau (x) \in \tau (X)$.