Problem
Given a Banach space $E$.
Regard a subspace: $$\iota:U\hookrightarrow E:u\mapsto u$$
Consider the projection: $$\pi:E'\twoheadrightarrow U':\psi\mapsto\psi\circ\iota$$
By Hahn-Banach find: $$\varphi\in U'\implies \varphi_E\in E'$$
It is a surjection: $$\pi[\varphi_E]=\varphi_E\circ\iota=\varphi$$
Also it is linear: $$\pi[\psi+\psi']=\pi[\psi]+\pi[\psi']\quad\pi[\lambda\psi]=\lambda\pi[\psi]$$
Then equivalence holds: $$E'\cong U'\iff E=\overline{U}$$
Especially one has then: $$\|\pi[\psi]\|=\|\psi\circ\iota\|=\|\psi\|$$
How can I check this?
Application
Consider the Banach space $\ell^\infty$.
Regard the closed spaces: $$\mathcal{c}_{(0)}=\overline{\mathcal{c}_{(0)}}<\ell^\infty$$
They have common duals: $$(\ell^\infty)',\mathcal{c}',\mathcal{c}_0'\cong\ell^1$$
Identifications can't be canonical!
Special thanks to Simon B.
By uniform extension: $$\varphi\in U'\implies\overline{\varphi}\in\overline{U}'$$
This happens uniquely: $$\overline{U}=E\implies\ker\pi=(0)$$
Especially one has then: $$\|\pi[\overline{\varphi}]\|=\|\overline{\varphi}\circ\iota\|=\|\varphi\|=\|\overline{\varphi}\|$$
Conversely assume: $$U=\overline{U}\neq E$$
By Riesz lemma: $$\|\hat{x}_0\|=1:\quad\|\hat{x}_0+U\|\geq\frac12$$
This gives rise to: $$\|u\|+\|\hat{x}_0\|\leq\|u+\hat{x}_0\|+2\|\hat{x}_0\|\leq5\|u+\hat{x}_0\|$$
Extend form distinctly: $$\varphi\in U':\quad\varphi_\pm(u+\lambda\hat{x}_0):=\varphi(u)\pm\|\varphi\|\|\hat{x}_0\|\lambda$$
They are both bounded: $$|\varphi_\pm(u+\lambda\hat{x}_0)|\leq|\varphi(u)|+\|\varphi\|\|\hat{x}_0\||\lambda|\\ \leq\|\varphi\|(\|u\|+\|\lambda\hat{x}_0\|)\leq5\|\varphi\|\cdot\|u+\lambda\hat{x}_0\|$$
Concluding the assertion.