You can apply the following statement to $X = \mathbb{P}^1_K$ and $L = O(1)$ when $K$ is a separably closed field.

Let $L$ be a line bundle on a reduced connected scheme $X$ such that $H^{0}(X,\mathcal{O}_X)$ is a separably closed field. Assume that any vector bundle on $X$ is of the form $\oplus_{i=1}^{n} L^{\otimes \lambda_i}$ for some integers $\lambda_1 \geq \dots \geq \lambda_n$, and that $L^{\lambda}$ has no nonzero global section for $\lambda < 0$. Then $X$ is simply connected.

Indeed, let $\mathcal{A}$ be a finite etale $\mathcal{O}_X$-algebra. Then $\mathcal{A} = \oplus_{i=1}^{n} L^{\otimes \lambda_i}$ as an $\mathcal{O}_X$-module for some integers $\lambda_1 \geq \dots \geq \lambda_n$. Because $\mathcal{A}$ has a global section (the unit) one must have $\lambda_1 \geq 0$.

If $\lambda_1 > 0$ then the morphism $L^{2 \lambda_1} \rightarrow \mathcal{A} \otimes_{\mathcal{O}_X} \mathcal{A} \rightarrow \mathcal{A} \rightarrow L^{\lambda_i}$ must be zero for each $i$ since $2 \lambda_1 > \lambda_i$. The the multiplication of $\mathcal{A}$ is $0$ on the $L^{ \lambda_1}$-factor, contradicting the reducedness of $\mathcal{A}$. Thus $\lambda_1 = 0$.

Since $\mathcal{A}$ is a finite etale $\mathcal{O}_X$-algebra the discriminant morphism
$$
\mathrm{det}(\mathcal{A})^{\otimes 2} \rightarrow \mathcal{O}_X
$$
is an isomorphism, so that $\sum_i \lambda_i =0$. Since $\lambda_{i} \leq 0$ for all $i$ we get that $\lambda_i = 0$ and thus that $\mathcal{A}$ is a trivial vector bundle. In particular the morphism
$$
H^0(X,\mathcal{A}) \otimes_{H^0(X,\mathcal{O}_X)} \mathcal{O}_X \rightarrow \mathcal{A}
$$
is an isomorphism. In particular $H^0(X,\mathcal{A})$ is a finite etale $H^0(X,\mathcal{O}_X)$-algebra, hence a trivial one, so that $\mathcal{A}$ is a trivial finite etale $\mathcal{O}_X$-algebra.