Shrink a banach space to a closed linear span of an into isometry

Question

Theorem $2.2$ states that

Let $X$ and $Y$ be separable Banach spaces and suppose that $f:X \rightarrow Y$ is an into isometry, then $X$ is linearly isometric to a subspace of $Y$.

In the proof, the author mentioned that

We may assume that $Y$ is equal to the closed linear span of $f(X)$.

Question: Why can we assume that?

Answer 1

If $X$ is isometric to a subspace of $\overline{f(X)}$, then of course it is isometric to a subspace of $Y$. So, we might as well assume $\overline{f(X)}=Y$.

Answer 2

Let $Z$ be the closed linear span of $f(X)$. Then, (due to the closure!) $Z$ is again a Banach space and $X$ is linearly isomorphic to a subspace of $Z$ iff it is linearly isomorphic to a subspacef of $Y$.