If $X$ is a variety over $k$, is it true that there exists a finite separable extension $k'$ of $k$ such that $X$ has a $k'$-point? What if we can assume $X$ is a smooth projective curve?
This seems silly, but I didn't know why couldn't all the residue fields at the closed points be inseparable over $k$.
This is true if you assume that $X/k$ is smooth. For convenience (since it's irrelevant), I'll assume that $X$ is equidimensional.
Let $x\in X$ be any point. By assumption that $X$ is smooth, there exists some neighborhood $U\subseteq X$ such that $X\to \text{Spec}(k)$ factors as
$$\begin{matrix}U & \xrightarrow{\text{etale}} & \mathbb{A}^n_k\\ & \searrow & \downarrow\\ & & \text{Spec}(k)\end{matrix}$$
where $n=\dim X$. Now, since $U\to X$ is etale, we know that its image $V$ is open. But, the $k$-points of $\mathbb{A}^n_k$ are dense. So, there exists $v\in V$ which is a $k$-point--let $u\in U$ be a point which maps to $v$. Then, by definition of etaleness, $\ell:=k(u)$ is a finite separable extension of $k(v)$. So, $u$ is an $\ell$-point with $\ell/k$ finite separable. Since $U(\text{Spec}(\ell))=U_\ell(\text{Spec}(\ell))$ we see that $U_\ell$ has an $ \ell$-point as desired.
As an exercise, try and extend the above to a much stronger statement about the separable points of $X$ (something to the tune of density).
If you don't assume that $X/k$ is smooth, then I believe the answer is no. Let $k=\overline{\mathbb{F}_p}(T)$, and let $X=\text{Spec}(k(T^{\frac{1}{p}})$. Suppose that $\ell/k$ is separable, and $X_\ell$ had an $\ell$-point. Well, $X_\ell=\text{Spec}(\ell[x]/(x^p-T)$. Then, $X_\ell$ having an $\ell$-point means that $x^p-T$ has a root in $\ell$. But, $x^p-T$ has a unique root in $\overline{k}$, and so clearly $\ell\supseteq k(T^{\frac{1}{p}})$, which implies that $\ell/k$ is not separable--contradiction.