$ Av=\lambda v \implies\ A^*v=\overline \lambda v $, Is this true only for normal operators? I know that is true for normal operators ( matrices). But isnt the following proof independant of normality.
$$\langle A^\ast v, v\rangle = \langle v, Av\rangle = \langle v, \lambda v\rangle = \langle \overline{\lambda} v,v\rangle $$
So we can conclude   $ A^\ast v=\overline{\lambda} $  , and we didnt use normality anywhere. 
If it is true that normality isnt needed for this then also normality isnt needed for orthogonal eigenvectors either. Where am i making a mistake ?

@Omnomnomnom is the proof of $ (AB)^\ast=B^\ast A^\ast $ that goes like this 
$$ \langle (AB) v, v\rangle= \langle v, (AB)^\ast v\rangle $$
$$ \langle (AB) v, v\rangle= \langle Bv, A^\ast v\rangle= \langle v, B^\ast A^\ast v\rangle$$   also wrong becuase we cant divide by vectors
                      in inner product
 A: Normality is essential. Note that $$\begin{align*}(A^*-\overline{\lambda})^*(A^*-\overline{\lambda})&=(A-\lambda)(A^*-\overline{\lambda})\\&=AA^*-\lambda A^*-\overline{\lambda}A+|\lambda|^2\\
&=A^*A-\lambda A^*-\overline{\lambda}A+|\lambda|^2\\
&=(A^*-\overline{\lambda})(A-\lambda)\\&=(A-\lambda)^*(A-\lambda).
\end{align*}$$
This shows that if $A$ is normal, then $A-\lambda$ is also normal. Since $\|Nx\|=\|N^*x\|$ for every normal operator $N$, it also holds that $\ker N = \ker N^*$. Thus
$$
\ker (A-\lambda) = \ker(A^*-\overline{\lambda})
$$ and it follows that
$$
Av=\lambda v\ \ \ \Longleftrightarrow \ \ \ A^*v = \overline{\lambda}v.
$$
EDIT (Necessity of normality, 1/3/2022)
To see that the normality of $A:V \to V$ is essential, we proceed by induction on the dimension of $V$.
If $\operatorname{dim}V = 1$, $A$ being normal is trivial.
Suppose $\operatorname {dim} V >1$. By the algebraic closedness of $\mathbb C$, we can find $w \ne 0$ and $\lambda \in \mathbb C$ such that $A w = \lambda w$. By the given condition, it is straightforward to see that $A^*w = \overline{\lambda} w$, and accordingly, $\langle w\rangle^\perp \equiv \left \{ v\in V: \langle v, w\rangle = 0\right\}$ is an $A$-invariant subspace. Now by the inductive hypothesis, $B \equiv A|_{\langle w\rangle^\perp} : \langle w\rangle^\perp \to \langle w\rangle^\perp$ is normal, which implies the normality of
$$
A = \left[
\begin{matrix}
\lambda & 0 \\
0 & B
\end{matrix}
\right].
$$
