I would like to know if the proof I have is correct.
Statement: Let $M$ be a closed subspace if a Banach space $X$. Let $\pi: X \rightarrow X/M$ be the quotient map. Put $Y= X/M$ for each $\varphi \, \in Y^{*}$ define $f \circ \varphi = \varphi\circ \pi$.
Prove that $f$ is well defined, linear isometry from $Y^*$ onto $M^{\perp}$
Attempted Proof:
(i) Well-defined: let $[x], \,[z] \in Y \, \mbox{s.t.} \, [x]=[z]$. Furthermore, let $\varphi \in Y^*$. Now since $[x]=[z]$ then $x+Y = z+Y \Rightarrow x-z \in Y$ and $\varphi(x) - \varphi(z) = \varphi (x-z) = 0 \Rightarrow \varphi(x) = \varphi(z)$ hence $f(\varphi(x)) = f(\varphi(z))$
(ii) Linearity: Let $\varphi, \, \psi \in Y^*\, \mbox{and} \, k \in \mathbb{C}$ then we have $f(k\varphi + \psi) = (k\varphi + \psi) \circ \pi = k\varphi\circ\pi + \psi\circ\pi = kf(\varphi) + f(\psi)$
(iii) Surjectivity (This is the one that seems the less correct): Let $x \in X$ and let $\varphi \in Y^*$ then $\pi(x) \in Y$ so $x\rightarrow \varphi(\pi(x))$ which is a continuous linear functional on $X$ and vanish if $x \in M$ so $f\circ \varphi \in M^{\perp}$
(iv) Isometry: Let $U$ be the unit ball in $X$ then $\pi(U) = U_{Y}$ is the unit ball in $Y$ , hence $\forall [x] \in U_Y \, \mbox{and} \, y \in [x]$ we have $\|f\circ\varphi([x])\| = \|\varphi\circ\pi(x)\| = \|\varphi\circ \pi(y)\| \leq \|\varphi\|\|y\| \Rightarrow \, |\varphi([x])| \leq \|\varphi\| \|[x]\| \, \mbox{hence} \, \|f\circ \varphi\| \leq \|\varphi\| $
Now to reverse the inequality,
$\forall x \in U_X \Rightarrow [x] \in U_Y$ which would imply that $\| f\circ \varphi\| \geq |f \circ \varphi([x])|= |\varphi(x)|$ so now we have $\| f\circ \varphi\| \geq \sup_{x\in U_X}|\varphi(x)| = \|\varphi\|$
so the inequalities are reversed hence isometric
Q.E.D.
You want to define a function $f:Y^*\rightarrow M^\perp$. That way, for $\varphi\in Y^*$ you must have an associated $f(\varphi)$ instead of $f\circ\varphi$. Also, in (i), you have $[x],[z]\in Y$, so $x,\in X$ and $\varphi\in Y^*$, so you cannot compute $\varphi(x)$ nor $\varphi(z)$ (actually, this step is not necessary). Surjectivity is not good at all. You simply showed that $f\circ\varphi\in M^\perp$. Step (iv) seems ok, but you should write $f\circ\varphi(x)$.
I would do the following: In step (i), define $f:Y^*\rightarrow M^\perp$ by $f(\varphi)=\varphi\circ\pi$. There are no arbitrary choices in the definition of $f$, so you don't have to show that it is well-defined. Your proof of it being an isometry os ok, just try to write it more carefully. Linearity should be obvious (if not, prove it).
Now surjectivity: Let $\Lambda\in M^\perp$. Define $\varphi\in Y^*$ by $\varphi([x])=\Lambda(x)$ for every $x\in X$. You have to show that $\varphi$ is well-defined, and it is linear and continuous. Obviously, $f(\varphi)=\Lambda$.