Yes. Let $p=char(K)$.
$K^{1/p}[x]\cap K[[x]]=K[x]$
Whence $K((x)) \cap K^{1/p}(x)= K(x)$ (if a rational function $u/v\in K^{1/p}(x)\cap K((x))$ then $v^p\in K[x]$ so that $u v^{p-1}\in K^{1/p}[x]\cap K[[x]]$)
Whence $$K((x)) \cap K(x)^{1/p}= K((x)) \cap K^{1/p}(x^{1/p})=K((x)) \cap K^{1/p}(x)= K(x)$$
Let $\alpha \in K((x))\cap \overline{K(x)}$
Its $K(x)$-minimal polynomial is $Q(y)\in K(x)[y]$. If it is inseparable then $Q(y) = R(y)^p$ with $$R(y)\in K(x)^{1/p}[y]\cap K((x))[y] = K(x)[y]$$
Which contradicts the minimality of $Q$.