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Choose a vector subspace $\mathfrak{m}$ complementary to the Lie algebra of a closed subgroup $H$. The differential of $F$ at $(0,0)$ is $(U,V)\mapsto U+V$, so the inverse function theorem makes $F$ a local diffeomorphism. Sufficiently small nonzero $U\in \mathfrak{m}$ cannot satisfy $\exp U\in H$: otherwise limit directions of a closed subgroup give a nonzero element of $\mathfrak{m}\cap \mathfrak{h}$. Consequently $H$ is locally exactly the coordinate slice $U=0$, proving that it is an embedded submanifold.

Suppose that $H$ is not a discrete subgroup. The identity is then an accumulation point of nonidentity subgroup elements: translate any failure of isolation to the identity. Choose $h_j\in H\setminus \{e\}$ with $h_j\to e$, and use a local exponential chart to write $h_j=\exp X_j$, where $X_j\ne 0$ and $X_j\to 0$ in the real Lie algebra $\mathfrak{g}$.

Choose a norm on this finite-dimensional vector space. By compactness of its unit sphere, pass to a subsequence with $X_j/\Vert X_j\Vert \to X$, where $\Vert X\Vert =1$. For an arbitrary real $t$, choose an integer $m_j$ nearest to $t/\Vert X_j\Vert$. Then $m_j\Vert X_j\Vert \to t$ andEvery left-hand element belongs to $H$, including for negative $m_j$. Closedness gives $\exp (tX)\in H$. Since $t$ was arbitrary and $X\ne 0$, the limit directions of a closed subgroup provide a nontrivial one-parameter subgroup in $H$. Its derivative at zero is $X$, so it cannot be the constant subgroup.

Choose a real vector subspace $\mathfrak{m}$ complementary to $\mathfrak{h}$, so $\mathfrak{g}=\mathfrak{m}\oplus \mathfrak{h}$. The smooth maphas 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 $H$.