The spectrum of self adjoint bounded operator on a Hilbert space - Why not use $\sigma(T^*)=\bar{\sigma(T)}$?

Question

Let $T$ be a bounded self-adjoint operator on a Hilbert space $\cal H$. Denote by $\sigma(T)$ the spectrum of $T$.
I wish to prove that the spectrum of a self-adjoint operator is contained in the real line.
I sense it is a dumb question, but is it not enough to use the fact that $\lambda\in \sigma(T)$ if and only if $\bar{\lambda}\in \sigma(T^*)$?

Since $T=T^*$, we'll have $\lambda=\bar{\lambda}$, thus $\lambda$ is real.
Note that I've encounter the several questions asked here regarding this issue (such as the ones I attached below), but none answered my question.

Spectrum of self-adjoint operator on Hilbert space real
On the spectrum of a self-adjoint operator

Answer 1

You can't deduce that $\lambda = \overline{\lambda}$ immediately. All you know is that if $\lambda \in \sigma(T)$, then there is $\mu \in \sigma(T)$ with $\lambda = \overline{\mu}$. How do you conclude $\lambda = \overline{\lambda}$ from this?