General Question
Let $X \hookrightarrow Y$ be an isometric embedding of a Banach space into another Banach space. Is it true that the bidual $X^{**}$ embeds isometrically into the bidual $Y^{**}$? It is clear that $X$ embeds in $Y^{**}$ via the canonical embedding but if $Y$ is not reflexive I see no reason why one should have $Y^{**} \subseteq X^{**}$.
More Specific Question
I'm reading this paper and in the proof of Lemma 3.7 the authors state the following. Let $G$ be a Hausdorff étale groupoid, $C^*(G)$ the maximal groupoid C*-algebra, and $\gamma \subseteq G$ an open bisection. Because $C_0(\gamma) \hookrightarrow C^*(G)$ is an isometric embedding of Banach spaces there are isometric embeddings $B^\infty(\gamma) \hookrightarrow C_0(\gamma)^{**} \hookrightarrow C^*(G)^{**}$ where $C^*(G)^{**}$ may be identified with the enveloping Von Neumann algebra and $B^\infty(\gamma)$ is the Banach space of bounded Borel functions on $\gamma$. I can see that the first inclusion is given by sending a Borel function $f$ to the linear functional on the dual of $C_0(\gamma)$ which integrates a measure against $f$ but I can't quite see the second inclusion.
Yes, $X^{**}$ embeds canonically in $Y^{**}$. Take the isometric embedding $\iota:X\to Y$. The double adjoint map $\iota^{**}:X^{**}\to Y^{**}$ is an isometric embedding, where $\iota^{**}(\chi)(\psi):=\chi(\psi\circ\iota)$ for all $\chi\in X^{**}$, $\psi\in Y^*$.
Note that every $\phi\in (X^*)_1$ is of the form $\psi\circ\iota$ for some $\psi\in (Y^*)_1$. Indeed, just consider the functional $\phi':\iota(X)\to\mathbb{C}$ given by $\phi'(\iota(x))=\phi(x)$ and extend it isometrically using Hahn-Banach to some $\psi:Y\to\mathbb{C}$. This justifies the following equality which yields isometricity of $\iota^{**}$. $$\|\iota^{**}(\chi)\|:=\sup_{\psi\in (Y^*)_1}|\chi(\psi\circ\iota)|=\sup_{\phi\in(X^*)_1}|\chi(\phi)|=:\|\chi\|_{X^{**}}.$$