In this question the OP tries to prove that only a real, even function $f(t)$ has a real Fourier transform $F(\omega)$.
Is it true? Is this the only case when the Fourier transform is real or are there other $f(t)$ that give rise to a real $F(\omega)$?
As Daniel Fischer said, the statement is false. The Fourier transform of $f$ is real-valued if and only if $f(-x) = \overline{f(x)}$ for almost all $x\in\mathbb R$.
To prove the necessity, apply the inverse Fourier transform to a real function $g(\xi)$: ignoring normalization constants, the computation goes as $$ \check g(-x) = \int e^{i(-x)\xi }g(\xi)\,d\xi = \int \overline{e^{ix\xi }}g(\xi)\,d\xi = \overline{\int e^{ix\xi } g(\xi)\,d\xi } = \overline{\check g(x)} $$
The proof of sufficiency is similar: if $f(-x) = \overline{f(x)}$ holds, then $$ \overline{\hat f(x)} = \overline{\int e^{-ix\xi} f(x)\,dx} =\int e^{ix\xi} f(-x)\,dx = \int e^{-ix\xi} f(x)\,dx = \hat f(x) $$ where the second-to-last step is the chang of variable, $x\mapsto -x$.