Fundamental group determined by connected components

This statement is nonsense. As you say, there is in fact no natural way to consider $\pi_0(G)$ as a subgroup of $\pi_0(H)$ if $H$ is a subgroup of $G$. Instead, there is a natural map $\pi_0(H)\to \pi_0(G)$ (though this map may not be injective).

What is true (assuming $H$ is a nice enough subgroup of $G$ that the quotient map $G\to G/H$ is a fibration) is that $\pi_1(G/H)$ is isomorphic to the kernel of this map $\pi_0(H)\to \pi_0(G)$. This follows easily from what you have already done, since a loop in $G/H$ lifts to a path in $G$ which then must be contained in the identity path component $G_0\subseteq G$, so $\pi_1(G/H)$ is the same as $\pi_1(G_0/(G_0\cap H))$. By what you have said, $\pi_1(G_0/(G_0\cap H))$ is naturally isomorphic to $\pi_0(G_0\cap H)$, which is exactly the kernel of $\pi_0(H)\to \pi_0(G)$.

[Incidentally, to me, a simply connected space must be path-connected, so $G$ must be path-connected, but your definitions may be different...]

As others have remarked, $\pi_0(G)$ is not naturally a subgroup of $\pi_0(H)$. The following example (which meets all of they hypothesis of your problem) indicates that sometimes it's not a subgroup at all.

Let $G = S^3\times \mathbb{Z}_2\times \mathbb{Z}_2$, where $S^3$ denotes the unit quaternions and $\mathbb{Z}_2 = \{\pm 1\}$ under multiplication. Note that $S^3$ is connected, so $\pi_0(G) = \mathbb{Z}_2\times \mathbb{Z}_2$. Note also that $G$ is simply connected (in the sense that each connected component is simply connected).

Let $H = \langle a,b\rangle$ where $a = (i,-1,1)$ and $b = (j,1,-1)$.

Claim 1: The group $H$ is isomorphic to the quaternion group $Q_8 =\{\pm 1, \pm i, \pm j, \pm k\}$, so that, in particular, $\pi_0(H)\cong Q_8$.

Proof: A simple calculation shows that all of the following relations hold: $$a^4 = e, \quad a^2 = b^2, \quad b^{-1}ab = a^{-1}.$$

According to wikipedia, one presentation of the Quaternion group is $\langle a,b: a^4 = e, a^2 = b^2, b^{-1} a b = a^{-1}\rangle$, so $H$ must be a quotient of $Q_8$. On the other hand, $\langle a\rangle$ contains $4$ elements, none of which are $b$, so $H$ consists of at least $5$ elements. Thus, $H$ is isomorphic to $Q_8$. $\square$

Claim 2: The space $G/H$ is connected.

Proof: Just note that $H$ meets all the components of $G$: the elements $a,a^2, b, ab$ are in different components. $\square$

Finally, simply note that $\pi_0(H)\cong H\cong Q_8$ has no subgroups isomorphic to $\pi_0(G)\cong \mathbb{Z}_2\times \mathbb{Z}_2$. In fact, all subgroups of order $4$ in $Q_8$ are cyclic.