Simply connected manifolds are orientable

For a simply connected $n$-manifold $M\subseteq\Bbb{R}^k$, I want to show that $M$ is orientable.

Take a point $p\in M$ and take an $n$-disc, $D^n$, around $p$ (we can take it as small as we please). Since $S^{n-1}$ is orientable and $M$ (and consequently $TM$) is simply connected, the orientation map $S^{n-1}\to TM$ can be extended to $D^n\to TM$. So around every point there is such a disc. We can construct an atlas (an orientation) out of these.

Does this suffice to prove the claim?

Let $M$ be a connected non-orientable manifold of dimension $n$ and let $M^*$ be its oriented double cover which is connected as $M$ is non-orientable. By general covering space theory, there is a short exact sequence $$0\to\pi_1(M^*)\to \pi_1(M)\stackrel{f}{\to}\mathbb{Z}/2\mathbb{Z}\stackrel{g}{\to} 0$$ and so if $\pi_1(M)$ is trivial, then $\mathbb{Z}/2\mathbb{Z} = \ker g =\operatorname{im}f = 0$ which is a contradiction.

Here is one way to argue. An orientation of a smooth $n$-manifold is a nowhere vanishing section of $\Lambda^n M$ (i.e., a volume form). This is a real line bundle $L\to M$. Put a Riemannian metric on $M$. This defines a metric on $L$ as well. Constructing a nowhere vanishing section of this bundle is the same as constructing a section of the unit sphere bundle $U\to M$ of $L$ (unit sphere in ${\mathbb R}$ is of course the set $\pm 1$). The unit sphere bundle $U\to M$ is a covering map (since the fiber is zero-dimensional). Since $M$ is simply-connected, the bundle $U\to M$ is trivial. Hence, it admits a section.

I realize that the question is old, but I wanted to present a more elementary approach, which only uses the fact that homotopically trivial loops lift to loops on a covering space.

Suppose $M$ is not orientable. Consider the oriented double cover $p:X \to M$ and pick $x_0 \in X$. Now, let $\lambda$ be a path connecting $x_0$ and the other point in the fiber, call it $x_1$ (here we are using the fact that the double cover of a non-orientable $M$ is connected - since it is a manifold, it is also path-connected). Now, $p \circ \lambda$ is a loop in $M$, which lifts to $\lambda$. Since $M$ is simply connected, $p \circ \lambda$ is homotopically trivial, which is an absurd (since $\lambda$ is not a loop).