Prove/Disprove an inner product on a complex linear space restricted to its real structure is also an inner product

This is not true in general. For instance, let $V=\mathbb{C}^2$. The vectors $v=(1,0)$ and $w=(i,1)$ are a basis for $V$, and there is a unique inner product $(\cdot,\cdot)$ on $V$ for which they are orthonormal. Now let $\sigma:V\to V$ be given by $\sigma(a,b)=(\overline{a},\overline{b})$. This is a conjugation, and $R_\sigma(V)=\mathbb{R}^2$. However, the inner product is not real-valued when restricted to $\mathbb{R}^2$: for instance, $v=(1,0)$ and $w-iv=(0,1)$ are both in $\mathbb{R}^2$, but $(v,w-iv)=i$ since $v$ and $w$ are orthonormal.

It is entirely possible that $(\alpha,\beta)\notin\Bbb R$ for some $\alpha,\beta\in R_\sigma(V)$. Since $\sigma$ is unrelated to $(\cdot,\cdot)$ and it only provides additional structure, but no restrictions, to the $\Bbb C$-linear structure of the space, you could take any $\Bbb C$-vector space $V$ of dimension $\ge2$ already endowed with a conjugate map $\sigma$, select a $\Bbb C$-basis $\{b_1,\cdots, b_n\}$ of $V$ made of vectors in $R_\sigma(V)$, and then endow $V$ with a positive-definite Hermitian product such that $(b_1,ib_2)=2$ (which you can do as freely as you can devise positive-definite Hermitian matrices).