= Solution
Choose a real <vector subspace> $\mathfrak m$ complementary to $\mathfrak h$, so $\mathfrak g=\mathfrak m\oplus\mathfrak h$. The smooth map
$$
F:\mathfrak m\times\mathfrak h\longrightarrow G,\qquad F(U,V)=\exp U\exp V
$$
has differential $(U,V)\mapsto U+V$ at $(0,0)$, an <isomorphism>. The <inverse function theorem> makes $F$ a <diffeomorphism> from a sufficiently small product neighborhood onto an open neighborhood of the identity in $G$.
There is a neighborhood of zero in $\mathfrak m$ in which $\exp U\in H$ implies $U=0$. Otherwise choose nonzero $U_j\in\mathfrak m$ tending to zero with $\exp U_j\in H$. A normalized subsequence tends to a unit vector $U\in\mathfrak m$. The <limit directions of a closed subgroup> from part (a) give $U\in\mathfrak h$, contradicting $\mathfrak m\cap\mathfrak h=0$. Shrink the product neighborhood so that this exclusion holds throughout its first factor.
If $F(U,V)\in H$, then $\exp V\in H$ by the definition of $\mathfrak h$, and hence $\exp U=F(U,V)\exp(-V)\in H$. The exclusion forces $U=0$. Conversely, every $F(0,V)=\exp V$ belongs to $H$. Thus the <local product coordinates for a closed subgroup> identify $H$ near the identity exactly with the smooth coordinate slice $U=0$.
Left translation by each element of $H$ gives the same description around that element, so $H$ is an <embedded submanifold> of $G$ in its subspace topology. The multiplication and inversion maps of $G$ restrict to smooth maps on this <embedded submanifold>, making $H$ a <Lie group> and an embedded <Lie subgroup>. Its tangent space at the identity is $\mathfrak h$, because $d\exp_0$ is the identity on this coordinate slice. It remains closed by hypothesis. This proves the <closed-subgroup theorem>, with no prior smoothness assumption on $H$.
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