I am unable to prove the following.
Let $E/F$ be a purely inseparable field extension (i.e. $E/F$ is an algebraic extension with F of characteristic $p$ and such that $\forall\ x \in E\; \exists\; n \geq 0\;x^{{p}^{n}}\in F$).
Let $X=\mathrm{Spec}{A}$ be an affine scheme over $F$.
Then the projection $\phi : X_{E} \to X$ is bijective.
In the notes, it is suggested that this can be done by showing that the scheme theoretic fiber of one point $\mathfrak{p} \in X$ is a single point. While I understand why this would work, I am not able to achieve this plan.
Indeed, I get $(X_{E})_{\mathfrak{p}}=\mathrm{Spec}{[(A \otimes_{F}E)\otimes_A({A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}})}]}$, from which I do not know how to conclude that this fiber is a single point.
This follows from the solution over at your other question and some work to compute the fiber. From the solution over there, we have that the fiber over $x\in X$ of $X_E\to X$ is given by the spectrum of the ring $k(x) \otimes_F E$. Let's calculate what this ring is.
As $E$ is a purely inseparable extension of $F$, it may be written as $E\cong F[z_i]/(z_i^{p^{n_i}}-c_i)$ for integers $n_i$ and elements $c_i\in F$ as $i$ ranges over some index set $I$. Therefore $k(x) \otimes_F E \cong k(x)[z_i]/(z_i^{p^{n_i}}-c_i)$, so every element in this ring is either algebraic over $k(x)$ or nilpotent, and therefore $k(x)\otimes_F E$ is a local ring of dimension zero with spectrum a single point.