For the first part, it's an important fact in itself that $X_{\overline{k}} \rightarrow X_{k^{sep}}$ is a homeomorphism.
To do that, it's not an issue to reduce to the case of affine $X$, and thus you want to show that for any $k$-algebra $A$, we have a natural correspondance between the spectra of $A \otimes \overline{k}$ and $A \otimes k^{sep}$. To do that, the main ingredient is to note that for each $x \in A \otimes \overline{k}$, in prime characteristic $p > 0$, there is an integer $n \geq 0$ such that $x^{p^n} \in A \otimes k^{sep}$. (It's not entirely obvious how to go on after that, but it's not very difficult).
Now, we want to explain why, if $X_{\overline{k}}$ is irreducible, then for any extension $k'/k$, $X_{k'}$ is irreducible.
First, we can assume $k$ is algebraically closed. Indeed, let $K$ be an algebraic closure of $k'$, so that we have a map $\overline{k} \rightarrow K$ (respecting $k' \cap \overline{k}$). Now $K/\overline{k}$ is a field extension, and $X_{\overline{k}}$ is irreducible. If $X_K$ is indeed irreducible, then, as $X_K \rightarrow X_{k'}$ is surjective, it follows that $X_{k'}$ will be irreducible.
So assume that $k$ is an algebraically closed field and $X$ is an irreducible scheme over $k$. Let $K/k$ be a field extension, we want to show that $X_{K}$ is irreducible. It's enough to show it when $X$ is affine (because the affine subsets of a general irreducible $X$ are dense and irreducible, and their base changes will be irreducible, and will pairwise intersect, which forces $X_K$ to be irreducible by topology), and we can furthermore assume $X$ is reduced (so $X$ is the spectrum of an integral $k$-algebra $A$), and we will show that if $K/k$ is an extension, $A\otimes K$ is an integral domain.
Let $K/k$ be an extension and let $a,b \in A \otimes K$ have a null product. There is a finitely generated (and integral) $k$-subalgebra $B \subset K$ such that $a,b$ are in the image of the monomorphism (flatness) $A \otimes_k B \rightarrow A \otimes K$, so we want to show that, in $A \otimes B$, $a$ or $b$ must be zero when their product is zero.
Now, take a $k$-basis $e_i$ of $A$, and let $a_i,b_i$ be the coordinates of $a,b$ in the $B$-basis $e_i \otimes 1$. Then, $(a_i)_i,(b_i)_i$ are almost null sequences of elements of $B$. Assume that there are some nonzero $a_i,b_j$. Then $a_ib_j \in B$ is nonzero (and $B$ is a finitely generated $k$-algebra), hence is not contained in some maximal ideal $\mu$ by the Nullstellensatz. Consider the reduction map mod $\mu$: $A \otimes B \rightarrow A$ (as $k$ is algebraically closed): then the images of neither $a$ nor $b$ are zero, but their product is, so $A$ isn't an integral domain, so we have a contradiction, which shows that $A \otimes K$ is a domain and we are done.
Best Answer
Unfortunately, your requested statements are not correct as written. The equivalence of categories can be fixed, but the "in other words" statement is false. Fortunately, there is a relatively quick and easy resolution to the question you ask at the end: any smooth variety which isn't geometrically irreducible splits over some finite extension in to a union of smooth geometrically irreducible varieties. So as long as you don't mind altering your base field up to a finite extension and you talk about the right sort of properties, you don't lose anything by assuming geometrically irreducible.
Let's handle the "in other words" statement first. If $X$ is smooth and birational to a geometrically irreducible variety, then $X$ must be geometrically irreducible, which is not exactly what you've written down - any smooth variety which is not geometrically irreducible is a counterexample to your statement. There are plenty such varieties: $\operatorname{Spec} \Bbb F_2[x]/(x^2+x+1)$, for instance, is smooth but not geometrically irreducible as over $\Bbb F_4$, it splits in to two smooth points.
To show the claim that smooth + birational to a geometrically irreducible variety implies geometrically irreducible, let $X$ be irreducible and smooth over $k$. First we note that any open subscheme of a geometrically irreducible subscheme is again geometrically irreducible, and geometric irreduciblity is preserved by isomorphisms. So by birationality, this means that there is an open subscheme $U$ of $X$ which is geometrically irreducible (ie $U_\overline{k}$ is irreducible). On the other hand, as $X$ is smooth over $k$, $X_\overline{k}$ is smooth over $\overline{k}$, so it's a disjoint union of irreducible components, each of which surjects on to $X$. In particular, the preimage of the generic point of $X$ is the collection of generic points of each irreducible component of $X_\overline{k}$. On the other hand, as $U$ is geometrically irreducible and has the same generic point as $X$, there can only be one of these generic points, so $X$ is geometrically irreducible.
Now let's get to the statement about categories. The easiest way to fix this up is to imitate one of the well-known equivalences between certain categories of curves over $k$ and fg transcendence degree one field extensions of $k$ and then add the condition about a algebraic closure of $k$ in this field from the comments to guarantee geometric irreducibility (the comments reference Liu's AGAC 3.2.14). For a reference on the equivalence of categories, see for instance Hartshorne I.6.12 or Stacks 0BY1. For instance, one possible correct statement would be that there's an equivalence of categories between smooth proper geometrically irreducible curves over $k$ and finitely generated field extensions $k\subset K$ of transcendence degree one with $k$ algebraically closed in $K$.