Q:

$M \leq X$. Is it true that $M^*$ can be isometrically embedded in $X^*$ where $*$ denotes the dual space?

Let $X$ be a Banach space and $M$ be a closed subspace i.e. $M \leq X$.
Is it true that $M^*$ can be isometrically embedded as a subspace of $X^*$ where $*$ denotes the dual space?
What I have tried:
I know by Hahn Banach theorem that every $f \in M^*$ has an extension $\hat f \in X^*$ such that $\hat f |_M =f$ and the operator norm doesn't change i.e. $\|\hat f\| =\|f\|$.
I found that such an extension can be chosen to be linear iff $X$ is a Hilbert space here, hence my claim might not hold in the general case.
Any help is appreciated.

A:

I have an assumption that I have not proved but which implies your conjecture.
Notation: For  $S\subset  X$ let $[ S]$ be the closed linear subspace of $X$ generated by  S. Let $\Bbb M$ be the set of closed linear subspaces  of $M.$ 
Let $I$ be the set of all $(Y,f)$ such that $Y\in \Bbb M $ and such that  $f:Y^*\to X^*$ is an isometric embedding satisfying  $\forall g\in Y^*\,(\,f(g)|_Y=g).$
For $(M_1,f_1),(M_2,f_2)\in I\, $  let $(M_1,f_1)\le (M_2,f_2)$ iff $M_1\subset M_2$ and $$\forall z\in (M_2)^*\,(\,f_2(z)|_{M_1}=f_1(z|_{M_1})|_{M_1}\,).$$
Unproven assumption: If $Y\ne M$ and $(Y,f)\in I$  there exists $x\in M$ \ $Y$ and $([Y\cup \{x\}],g)\in I$ with $(Y,f)\le ([Y\cup \{x\}],g).$
Suppose $C=\{(M_j,f_j):j\in J\}\,$ is any non-empty $\le$-chain in $I.$  For $z\in [\cup_{j\in J}M_j]^*$ and $k\in J\,$   let $$f(z)|_{M_k} =f_k(z|_{M_k})|_{M_k}.$$ Then each $f(z)$ is a bounded linear function on $\cup_{j\in J}M_j$ so it extends uniquely to $g(z)\in [\cup_{j\in J}M_j]^*.\, $ And by  the Hahn-Banach Theorem $g(z)$ extends to $h(z)\in X^*$ with $\|g(z)\|=\|h(z)\|.$  Then $([\cup_{j\in J} M_j],h)$ is a $\le$-upper bound for $C.$
So every non-empty $\le$-chain has a $\le$-upper bound.
Now $\le$ is transitive and reflexive. And $I$ is not empty, for if $Y=\{0\}$ and $f(0)=0\in X^*$ then $(Y,f)\in I. $
So by Zorn's Lemma, $I$ has a $\le$-maximal member $(Y',f').$ My unproven assumption implies that $Y'=M.$ 

