Let $(M,g)$ be a connected (not necessarily complete) Riemannian manifold with constant sectional curvature. Then is $(M,g)$ isometric to an open subset of a space form?
(By space form, I mean a complete connected Riemannian manifold of constant sectional curvature.)
Not necessarily. Here's an interesting counterexample. Let $N$ be $\mathbb S^2$ with the north and south poles removed, and let $M$ be the universal cover of $N$. (Since $N$ is diffeomorphic to $\mathbb R \times \mathbb S^1$, $M$ is diffeomorphic to $\mathbb R^2$.) Give $M$ the constant-curvature metric obtained by pulling back the round metric from $N$ via the covering map. (You can visualize the covering map as an infinitely long orange peel wrapping around the sphere countably many times.)
Suppose $S$ is a space form and $F\colon M\to S$ is an isometric embedding. There's a Riemannian covering $\pi\colon \mathbb S^2 \to S$. Since $M$ is simply connected, it follows from covering space theory (see [ITM, Thm. 11.18]) that $F$ has a lift to a continuous map $\widetilde F\colon M\to \mathbb S^2$, and since $F$ and $\pi$ are local isometries, so is $\widetilde F$. Moreover, the fact that $F$ is injective implies $\widetilde F$ is also injective.
Now let $\gamma\colon \mathbb R\to \mathbb S^2$ be a unit-speed periodic geodesic whose image is the equator in $\mathbb S^2$, and let $\widetilde \gamma \colon\mathbb R \to M$ be the geodesic that projects down to the equator under the covering map. Then $\widetilde F \circ \widetilde \gamma$ is an infinitely long injective geodesic in $\mathbb S^2$, which is impossible.
[ITM] John M. Lee, Introduction to Topological Manifolds, 2nd ed.