Let $T \in \mathcal{L}(V)$ where $V \subseteq \mathbb{C}^n$. Furthermore, let $v, w \in V$ such that $v$ is an eigenvector of $T$ corresponding to eigenvalue $\lambda$ (i.e. $Tv = \lambda v$). It follows that
$$ \left<Tv, w\right> = \left<\lambda v, w\right> = \left<v, \overline{\lambda}w\right> $$
since $\lambda$ is an eigenvalue of $T$ if and only if $\overline{\lambda}$ is an eigenvalue of $T^*$. Furthermore, by the definition of adjoint operators,
$$ \left<Tv, w\right> = \left<v, T^*w\right> $$ These two equalities imply that
$$ \left< v, T^*w\right> = \left<v, \overline{\lambda}w\right> $$
for any choice of $w \in V$. This implies that any vector in $V$ is an eigenvector of $T^*$ corresponding to eigenvalue $\overline{\lambda}$. However, this is clearly not true. Could somebody please explain this?
The error was pointed out in the comments by John and Mitchell Faulk. Although it's true that
$$\left(\forall v\in V \ \langle v, T^*w\rangle = \left\langle v, \overline{\lambda}w\right\rangle \right)\implies T^*w = \bar \lambda w $$
in the post the equality on the left was shown only for a specific $v$. This does not imply $T^*w = \bar \lambda w$, but only that $T^*w - \bar \lambda w$ is orthogonal to $v$.