Let $k$ be a field (not necessarily algebraically closed). We call $k$-variety a scheme of finite type over $\mathrm{Spec}\left(k\right)$. Let $X$ be a geometrically reduced $k$-variety and $Y$ a $k$-variety.
The rest of the problem statement is as follows: Let $f,g:X\rightarrow Y$ two $k$-morphisms. We suppose that the set-wise applications $X\left(\bar{k}\right)\rightarrow Y\left(\bar{k}\right)$ induced by $f$ and $g$ coincide. We wish to show that $f=g$.
Question: How can it be shown that we can suppose $X$ and $Y$ to be affine?
Let $X$ be a geometrically reduced $k$-variety and $Y$ a $k$-variety. Let $(Y_i)$ be an open affine covering of $Y$. Then $f^{-1}(Y_i)$ is an open (by continuity of $f$) covering of $X$, but this covering is not affine in general as $f$ or $g$ might not be quasi-affine morphisms. Nevertheless take an open affine cover $V_{ij, f}$ of $f^{-1}(Y_i)$. The open $V_{ij, f} \cap V_{ik, g}$ are affine, this is implied by the stability of open immersions by base change, see here. Now look at $ f,g : V_{ij, f} \cap V_{ik, g} \rightarrow Y_i$ : these restrictions of $f,g$ verify the hypothesis (exercise, check it !) and are affine so they are equal. From this follows the conclusion.