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Then $\mathfrak{h}$ is invariant under the action by conjugation of $\mathrm{SO}_n(\mathbf{R})$. Now (see the post by Venkataramana), we can decompose $\mathfrak{sl}_n(\mathbf{R})$ as a $\mathrm{SO}_n(\mathbf{R})$-representation, namely $\mathfrak{gl}_n(\mathbf{R}) = \mathfrak{so}_n(\mathbf{R})\oplus\mathfrak{p}$, where $\mathfrak{p}$ is the set of symmetric matrices with trace zero. Then for all $n \ge 2$ (including $n=2$), $\mathfrak{p}$ is irreducible; it follows that the subspaces of $\mathfrak{gl}_n(\mathbf{R})$ containing $\mathfrak{so}_n(\mathbf{R})$ and invariant under conjugation by $\mathrm{SO}_n(\mathbf{R})$ are $\mathfrak{so}_n(\mathbf{R})$ and $\mathfrak{so}_n(\mathbf{R})\oplus\mathfrak{p}=\mathfrak{sl}_n(\mathbf{R})$.

If $H$ is contained in $\mathrm{SO}_n(\mathbf{R})$ then we are in one of the first cases. Conversely, if $H$ is not contained in $\mathrm{SO}_n(\mathbf{R})$ then since the latter is the stabilizer of $\mathfrak{so}_n(\mathbf{R})$ in $\mathrm{SL}_n(\mathbf{R})$, we deduce that $\mathfrak{h}$ contains a conjugate of $\mathfrak{so}_n(\mathbf{R})$ distinct from $\mathfrak{so}_n$. Hence we are in the second case, that is, $\mathfrak{h}=\mathfrak{sl}_n(\mathbf{R})$. In this case we can pick a basis $(e_1,\dots,e_m)$ of $\mathfrak{sl}_n(\mathbf{R})$ ($m=n^2-1$) and functions $u_1,\dots,u_m:\mathbf{R}\to H$ as in the definition of $\mathfrak{h}$ with $u'_i(0)=e_i$. It follow that the differential of the function $(u_1,\dots,u_m):\mathbf{R}^m\to H\subset\mathrm{SL}_n(\mathbf{R})$ at zero is surjective. Hence its image contains a neighborhood of $I_n$ in $\mathrm{SL}_n(\mathbf{R})$. Thus $H$ is open in $\mathrm{SL}_n(\mathbf{R})$ and hence by connectedness of the latter, is equal to $\mathrm{SL}_n(\mathbf{R})$.

Now remove the assumption that $H$ is in $\mathrm{SL}_n(\mathbf{R})$. If $H\cap\mathrm{SL}_n(\mathbf{R})=\mathrm{SO}_n(\mathbf{R})$, then $H$ is contained in the normalizer $\mathbf{R}^*\mathrm{O}_n(\mathbf{R})$ of $\mathrm{SO}_n(\mathbf{R})$. Otherwise, we deduce from the previous case that $H\cap\mathrm{SL}_n(\mathbf{R})$ equals $\mathrm{SL}_n(\mathbf{R})$, that is to say, $H$ contains $\mathrm{SL}_n(\mathbf{R})$. This finishes the proof.

Then $\mathfrak{h}$ is invariant under the action by conjugation of $\mathrm{SO}_n(\mathbf{R})$. Now (as mentioned in the post by Venkataramana), we can decompose $\mathfrak{sl}_n(\mathbf{R})$ as a $\mathrm{SO}_n(\mathbf{R})$-representation, namely $\mathfrak{sl}_n(\mathbf{R}) = \mathfrak{so}_n(\mathbf{R})\oplus\mathfrak{p}$, where $\mathfrak{p}$ is the set of symmetric matrices with trace zero. Then for all $n \ge 2$ (including $n=2$), $\mathfrak{p}$ is irreducible (see a proof below); it follows that the subspaces of $\mathfrak{sl}_n(\mathbf{R})$ containing $\mathfrak{so}_n(\mathbf{R})$ and invariant under conjugation by $\mathrm{SO}_n(\mathbf{R})$ are $\mathfrak{so}_n(\mathbf{R})$ and $\mathfrak{so}_n(\mathbf{R})\oplus\mathfrak{p}=\mathfrak{sl}_n(\mathbf{R})$.

If $H$ is contained in $\mathrm{SO}_n(\mathbf{R})$ then we are in one of the first cases. Conversely, if $H$ is not contained in $\mathrm{SO}_n(\mathbf{R})$ then since the latter is the stabilizer of $\mathfrak{so}_n(\mathbf{R})$ in $\mathrm{SL}_n(\mathbf{R})$ (see a proof below), we deduce that $\mathfrak{h}$ contains a conjugate of $\mathfrak{so}_n(\mathbf{R})$ distinct from $\mathfrak{so}_n$. Hence we are in the second case, that is, $\mathfrak{h}=\mathfrak{sl}_n(\mathbf{R})$. In this case we can pick a basis $(e_1,\dots,e_m)$ of $\mathfrak{sl}_n(\mathbf{R})$ ($m=n^2-1$) and functions $u_1,\dots,u_m:\mathbf{R}\to H$ as in the definition of $\mathfrak{h}$ with $u'_i(0)=e_i$. It follow that the differential of the function $(u_1,\dots,u_m):\mathbf{R}^m\to H\subset\mathrm{SL}_n(\mathbf{R})$ at zero is surjective. Hence its image contains a neighborhood of $I_n$ in $\mathrm{SL}_n(\mathbf{R})$. Thus $H$ is open in $\mathrm{SL}_n(\mathbf{R})$ and hence by connectedness of the latter, is equal to $\mathrm{SL}_n(\mathbf{R})$.

Now remove the assumption that $H$ is in $\mathrm{SL}_n(\mathbf{R})$. If $H\cap\mathrm{SL}_n(\mathbf{R})=\mathrm{SO}_n(\mathbf{R})$, then $H$ is contained in the normalizer $\mathbf{R}^*\mathrm{O}_n(\mathbf{R})$ of $\mathrm{SO}_n(\mathbf{R})$. Otherwise, we deduce from the previous case that $H\cap\mathrm{SL}_n(\mathbf{R})$ equals $\mathrm{SL}_n(\mathbf{R})$, that is to say, $H$ contains $\mathrm{SL}_n(\mathbf{R})$. This finishes the proof.

Addendum: proof of some basic facts (asked in a comment)

Proof that the action of $\mathrm{SO}_n(\mathbf{R})$ on $\mathfrak{p}$ is irreducible for $n\ge 2$. First, consider the action of the subgroup of signed permutation matrices with determinant 1 on the subspace $D\subset\mathfrak{p}$ of diagonal matrices with trace zero. It factors through the action of permutation matrices (not only even ones, using $n\ge 2$) on the set of $n$-vectors with sum zero (by permutation of coordinates), and this action is well-known and easily checked to be irreducible.

Now let $V$ be a nonzero $\mathrm{SO}_n(\mathbf{R})$-invariant subspace of $\mathfrak{p}$. Picking a nonzero element in $V$, it has a $\mathrm{SO}_n(\mathbf{R})$-conjugate that is diagonal. Therefore $V\cap D\neq 0$, and by the previous irreducibility fact, $D\subset V$. Now any element of $\mathfrak{p}$ has a $\mathrm{SO}_n(\mathbf{R})$-conjugate that is diagonal, so $V=\mathfrak{p}$. $\Box$

Proof that the stabilizer $N$ of $\mathfrak{so}_n(\mathbf{R})$ in $\mathrm{SL}_n(\mathbf{R})$ is $\mathrm{SO}_n(\mathbf{R})$. The set of $\mathfrak{so}_n(\mathbf{R})$-invariant scalar products on $\mathbf{R}^n$ is the line generated by the standard scalar product (that it is reduced to a line follows from the absolute irreducibility of the $\mathfrak{so}_n(\mathbf{R})$-action on $\mathbf{R}^n$). Hence this line is $N$-invariant. The stabilizer (in $\mathrm{GL}_n(\mathbf{R})$) of this line is by definition the group of similarities, hence its stabilizer in $\mathrm{SL}_n(\mathbf{R})$ is the group of similarities of determinant 1, that is, $N=\mathrm{SO}_n(\mathbf{R})$. $\Box$

Here's a complete proof of:

Every subgroup of $\mathrm{GL}_n(\mathbf{R})$ containing $\mathrm{SO}_n(\mathbf{R})$ is either contained in the group of similarities $\mathbf{R}^*\mathrm{O}(n)$, or contains $\mathrm{SL}_n(\mathbf{R})$.

(This is equivalent to the statement that for every $t\in\mathrm{GL}_n(\mathbf{R})\smallsetminus \mathbf{R}^*\mathrm{O}(n)$, the subgroup generated by $\mathrm{SO}_n(\mathbf{R})\cup\{t\}$ contains $\mathrm{SL}_n(\mathbf{R})$. Q2 for $n\ge 2$ and Q1 for $n\ge 3$ immediately follow.)

Let $H$ be a subgroup of $\mathrm{GL}_n(\mathbf{R})$. Let $\mathfrak{h}$ be the set of $x\in\mathfrak{gl}_n(\mathbf{R})$ such that there exists a $C^1$ map $u:\mathbf{R}\to\mathrm{GL}_n(\mathbf{R})$ such that $u(0)=I_n$ (the identity matrix), $u'(0)=x$, and $u$ takes values in $H$. Then $\mathfrak{h}$ is a linear subspace. Indeed, it's clearly stable under scalar multiplication, and if $u,v$ are two such maps with $u'(0)=x$, $v'(0)=y$, then defining $w(t)=u(t)v(t)$ we have $w'(0)=x+y$.

Now assume that $H$ contains $\mathrm{SO}(n)$.

First assume that $H$ is contained in $\mathrm{SL}_n(\mathbf{R})$; it follows that $\mathfrak{h}$ is contained in $\mathfrak{sl}_n$.

Then $\mathfrak{h}$ is invariant under the action by conjugation of $\mathrm{SO}(n)$. Now (see the post by Venkataramana), we can decompose $\mathfrak{sl}_n(\mathbf{R})$ as a $\mathrm{SO}(n)$-representation, namely $\mathfrak{gl}_n(\mathbf{R})=\mathfrak{so}_n\oplus\mathfrak{p}$, where $\mathfrak{p}$ are the symmetric matrices with trace zero. Then for all $n\ge 2$ (including $n=2$), $\mathfrak{p}$ is irreducible; it follows that the subspaces of $\mathfrak{gl}_n$ containing $\mathfrak{so}_n$ and invariant under conjugation by $\mathrm{SO}(n)$ are $\mathfrak{so}_n$ are $\mathfrak{so}_n\oplus\mathfrak{p}=\mathfrak{sl}_n$.

If $H$ is contained in $\mathrm{SO}(n)$ then we are in one of the first case. Conversely, if $H$ is not contained in $\mathrm{SO}(n)$ then since the latter is the stabilizer of $\mathfrak{so}_n$ in $\mathrm{SL}_n(\mathbf{R})$, we deduce that $\mathfrak{h}$ contains a conjugate of $\mathfrak{so}_n$ distinct from $\mathfrak{so}_n$. Hence we are in the second case, that is, $\mathfrak{h}=\mathfrak{sl}_n$. In this case we can pick a basis $(e_1,\dots,e_m)$ of $\mathfrak{sl}_n$ ($m=n^2-1$) and functions $u_1,\dots,u_m:\mathbf{R}\to H$ as in the definition of $\mathfrak{h}$ with $u'_i(0)=e_i$. It follow that the differential of the function $(u_1,\dots,u_m):\mathbf{R}^m\to H\subset\mathrm{SL}_n(\mathbf{R})$ at zero is surjective. Hence its image contains a neighborhood of $I_n$ in $\mathrm{SL}_n(\mathbf{R})$. Thus $H$ is open in $\mathrm{SL}_n(\mathbf{R})$ and hence by connectedness of the latter, is equal to $\mathrm{SL}_n(\mathbf{R})$.

Now remove the assumption that $H$ is in $\mathrm{SL}_n$. If $H\cap\mathrm{SL}_n=\mathrm{SO}_n$, then $H$ is contained in the normalizer $\mathbf{R}^*\mathrm{O}_n(\mathbf{R})$ of $\mathrm{SO}_n$. Otherwise, we deduce from the previous case that $H\cap\mathrm{SL}_n$ equals $\mathrm{SL}_n$, that is to say, $H$ contains $\mathrm{SL}_n$. This finishes the proof.

Here's a complete proof of:

Every subgroup of $\mathrm{GL}_n(\mathbf{R})$ containing $\mathrm{SO}_n(\mathbf{R})$ is either contained in the group of similarities $\mathbf{R}^*\mathrm{O}_n(\mathbf{R})$, or contains $\mathrm{SL}_n(\mathbf{R})$.

(This is equivalent to the statement that for every $t\in\mathrm{GL}_n(\mathbf{R})\smallsetminus \mathbf{R}^*\mathrm{O}(n)$, the subgroup generated by $\mathrm{SO}_n(\mathbf{R})\cup\{t\}$ contains $\mathrm{SL}_n(\mathbf{R})$. The answers to Q2 for $n\ge 2$ and to Q1 for $n\ge 3$ immediately follow.)

Let $H$ be a subgroup of $\mathrm{GL}_n(\mathbf{R})$. Let $\mathfrak{h}$ be the set of $x\in\mathfrak{gl}_n(\mathbf{R})$ such that there exists a $C^1$ map $u:\mathbf{R}\to\mathrm{GL}_n(\mathbf{R})$ such that $u(0)=I_n$ (the identity matrix), $u'(0)=x$, and $u$ takes values in $H$. Then $\mathfrak{h}$ is a linear subspace. Indeed, it's clearly stable under scalar multiplication, and if $u,v$ are two such maps with $u'(0)=x$, $v'(0)=y$, then defining $w(t)=u(t)v(t)$ we have $w'(0)=x+y$.

Now assume that $H$ contains $\mathrm{SO}_n(\mathbf{R})$.

First assume that $H$ is contained in $\mathrm{SL}_n(\mathbf{R})$; it follows that $\mathfrak{h}$ is contained in $\mathfrak{sl}_n(\mathbf{R})$.

Then $\mathfrak{h}$ is invariant under the action by conjugation of $\mathrm{SO}_n(\mathbf{R})$. Now (see the post by Venkataramana), we can decompose $\mathfrak{sl}_n(\mathbf{R})$ as a $\mathrm{SO}_n(\mathbf{R})$-representation, namely $\mathfrak{gl}_n(\mathbf{R}) = \mathfrak{so}_n(\mathbf{R})\oplus\mathfrak{p}$, where $\mathfrak{p}$ is the set of symmetric matrices with trace zero. Then for all $n \ge 2$ (including $n=2$), $\mathfrak{p}$ is irreducible; it follows that the subspaces of $\mathfrak{gl}_n(\mathbf{R})$ containing $\mathfrak{so}_n(\mathbf{R})$ and invariant under conjugation by $\mathrm{SO}_n(\mathbf{R})$ are $\mathfrak{so}_n(\mathbf{R})$ and $\mathfrak{so}_n(\mathbf{R})\oplus\mathfrak{p}=\mathfrak{sl}_n(\mathbf{R})$.

If $H$ is contained in $\mathrm{SO}_n(\mathbf{R})$ then we are in one of the first cases. Conversely, if $H$ is not contained in $\mathrm{SO}_n(\mathbf{R})$ then since the latter is the stabilizer of $\mathfrak{so}_n(\mathbf{R})$ in $\mathrm{SL}_n(\mathbf{R})$, we deduce that $\mathfrak{h}$ contains a conjugate of $\mathfrak{so}_n(\mathbf{R})$ distinct from $\mathfrak{so}_n$. Hence we are in the second case, that is, $\mathfrak{h}=\mathfrak{sl}_n(\mathbf{R})$. In this case we can pick a basis $(e_1,\dots,e_m)$ of $\mathfrak{sl}_n(\mathbf{R})$ ($m=n^2-1$) and functions $u_1,\dots,u_m:\mathbf{R}\to H$ as in the definition of $\mathfrak{h}$ with $u'_i(0)=e_i$. It follow that the differential of the function $(u_1,\dots,u_m):\mathbf{R}^m\to H\subset\mathrm{SL}_n(\mathbf{R})$ at zero is surjective. Hence its image contains a neighborhood of $I_n$ in $\mathrm{SL}_n(\mathbf{R})$. Thus $H$ is open in $\mathrm{SL}_n(\mathbf{R})$ and hence by connectedness of the latter, is equal to $\mathrm{SL}_n(\mathbf{R})$.

Now remove the assumption that $H$ is in $\mathrm{SL}_n(\mathbf{R})$. If $H\cap\mathrm{SL}_n(\mathbf{R})=\mathrm{SO}_n(\mathbf{R})$, then $H$ is contained in the normalizer $\mathbf{R}^*\mathrm{O}_n(\mathbf{R})$ of $\mathrm{SO}_n(\mathbf{R})$. Otherwise, we deduce from the previous case that $H\cap\mathrm{SL}_n(\mathbf{R})$ equals $\mathrm{SL}_n(\mathbf{R})$, that is to say, $H$ contains $\mathrm{SL}_n(\mathbf{R})$. This finishes the proof.

Here's a complete proof of:

Every subgroup of $\mathrm{GL}_n(\mathbf{R})$ containing $\mathrm{SO}_n(\mathbf{R})$ is either contained in the group of similarities $\mathbf{R}^*\mathrm{O}(n)$, or contains $\mathrm{SL}_n(\mathbf{R})$.

(This is equivalent to the statement that for every $t\in\mathrm{GL}_n(\mathbf{R})\smallsetminus \mathbf{R}^*\mathrm{O}(n)$, the subgroup generated by $\mathrm{SO}_n(\mathbf{R})\cup\{t\}$ contains $\mathrm{SL}_n(\mathbf{R})$. Q2 for $n\ge 2$ and Q1 for $n\ge 3$ immediately follow.)

Let $H$ be a subgroup of $\mathrm{GL}_n(\mathbf{R})$. Let $\mathfrak{h}$ be the set of $x\in\mathfrak{gl}_n(\mathbf{R})$ such that there exists a $C^1$ map $u:\mathbf{R}\to\mathrm{GL}_n(\mathbf{R})$ such that $u(0)=I_n$ (the identity matrix) and $u'(0)=x$. Then $\mathfrak{h}$ is a linear subspace. Indeed, it's clearly stable under scalar multiplication, and if $u,v$ are two such maps with $u'(0)=x$, $v'(0)=y$, then defining $w(t)=u(t)v(t)$ we have $w'(0)=x+y$.

Now assume that $H$ contains $\mathrm{SO}(n)$.

First assume that $H$ is contained in $\mathrm{SL}_n(\mathbf{R})$; it follows that $\mathfrak{h}$ is contained in $\mathfrak{sl}_n$.

Then $\mathfrak{h}$ is invariant under the action by conjugation of $\mathrm{SO}(n)$. Now (see the post by Venkataramana), we can decompose $\mathfrak{sl}_n(\mathbf{R})$ as a $\mathrm{SO}(n)$-representation, namely $\mathfrak{gl}_n(\mathbf{R})=\mathfrak{so}_n\oplus\mathfrak{p}$, where $\mathfrak{p}$ are the symmetric matrices with trace zero. Then for all $n\ge 2$ (including $n=2$), $\mathfrak{p}$ is irreducible; it follows that the subspaces of $\mathfrak{gl}_n$ containing $\mathfrak{so}_n$ and invariant under conjugation by $\mathrm{SO}(n)$ are $\mathfrak{so}_n$ are $\mathfrak{so}_n\oplus\mathfrak{p}=\mathfrak{sl}_n$.

If $H$ is contained in $\mathrm{SO}(n)$ then we are in one of the first case. Conversely, if $H$ is not contained in $\mathrm{SO}(n)$ then since the latter is the stabilizer of $\mathfrak{so}_n$ in $\mathrm{SL}_n(\mathbf{R})$, we deduce that $\mathfrak{h}$ contains a conjugate of $\mathfrak{so}_n$ distinct from $\mathfrak{so}_n$. Hence we are in the second case, that is, $\mathfrak{h}=\mathfrak{sl}_n$. In this case we can pick a basis $(e_1,\dots,e_m)$ of $\mathfrak{sl}_n$ ($m=n^2-1$) and functions $u_1,\dots,u_m:\mathbf{R}\to H$ as in the definition of $\mathfrak{h}$ with $u'_i(0)=e_i$. It follow that the differential of the function $(u_1,\dots,u_m):\mathbf{R}^m\to H\subset\mathrm{SL}_n(\mathbf{R})$ at zero is surjective. Hence its image contains a neighborhood of $I_n$ in $\mathrm{SL}_n(\mathbf{R})$. Thus $H$ is open in $\mathrm{SL}_n(\mathbf{R})$ and hence by connectedness of the latter, is equal to $\mathrm{SL}_n(\mathbf{R})$.

Now remove the assumption that $H$ is in $\mathrm{SL}_n$. If $H\cap\mathrm{SL}_n=\mathrm{SO}_n$, then $H$ is contained in the normalizer $\mathbf{R}^*\mathrm{O}_n(\mathbf{R})$ of $\mathrm{SO}_n$. Otherwise, we deduce from the previous case that $H\cap\mathrm{SL}_n$ equals $\mathrm{SL}_n$, that is to say, $H$ contains $\mathrm{SL}_n$. This finishes the proof.

Here's a complete proof of:

Every subgroup of $\mathrm{GL}_n(\mathbf{R})$ containing $\mathrm{SO}_n(\mathbf{R})$ is either contained in the group of similarities $\mathbf{R}^*\mathrm{O}(n)$, or contains $\mathrm{SL}_n(\mathbf{R})$.

(This is equivalent to the statement that for every $t\in\mathrm{GL}_n(\mathbf{R})\smallsetminus \mathbf{R}^*\mathrm{O}(n)$, the subgroup generated by $\mathrm{SO}_n(\mathbf{R})\cup\{t\}$ contains $\mathrm{SL}_n(\mathbf{R})$. Q2 for $n\ge 2$ and Q1 for $n\ge 3$ immediately follow.)

Let $H$ be a subgroup of $\mathrm{GL}_n(\mathbf{R})$. Let $\mathfrak{h}$ be the set of $x\in\mathfrak{gl}_n(\mathbf{R})$ such that there exists a $C^1$ map $u:\mathbf{R}\to\mathrm{GL}_n(\mathbf{R})$ such that $u(0)=I_n$ (the identity matrix), $u'(0)=x$, and $u$ takes values in $H$. Then $\mathfrak{h}$ is a linear subspace. Indeed, it's clearly stable under scalar multiplication, and if $u,v$ are two such maps with $u'(0)=x$, $v'(0)=y$, then defining $w(t)=u(t)v(t)$ we have $w'(0)=x+y$.

Now assume that $H$ contains $\mathrm{SO}(n)$.

First assume that $H$ is contained in $\mathrm{SL}_n(\mathbf{R})$; it follows that $\mathfrak{h}$ is contained in $\mathfrak{sl}_n$.

Then $\mathfrak{h}$ is invariant under the action by conjugation of $\mathrm{SO}(n)$. Now (see the post by Venkataramana), we can decompose $\mathfrak{sl}_n(\mathbf{R})$ as a $\mathrm{SO}(n)$-representation, namely $\mathfrak{gl}_n(\mathbf{R})=\mathfrak{so}_n\oplus\mathfrak{p}$, where $\mathfrak{p}$ are the symmetric matrices with trace zero. Then for all $n\ge 2$ (including $n=2$), $\mathfrak{p}$ is irreducible; it follows that the subspaces of $\mathfrak{gl}_n$ containing $\mathfrak{so}_n$ and invariant under conjugation by $\mathrm{SO}(n)$ are $\mathfrak{so}_n$ are $\mathfrak{so}_n\oplus\mathfrak{p}=\mathfrak{sl}_n$.

If $H$ is contained in $\mathrm{SO}(n)$ then we are in one of the first case. Conversely, if $H$ is not contained in $\mathrm{SO}(n)$ then since the latter is the stabilizer of $\mathfrak{so}_n$ in $\mathrm{SL}_n(\mathbf{R})$, we deduce that $\mathfrak{h}$ contains a conjugate of $\mathfrak{so}_n$ distinct from $\mathfrak{so}_n$. Hence we are in the second case, that is, $\mathfrak{h}=\mathfrak{sl}_n$. In this case we can pick a basis $(e_1,\dots,e_m)$ of $\mathfrak{sl}_n$ ($m=n^2-1$) and functions $u_1,\dots,u_m:\mathbf{R}\to H$ as in the definition of $\mathfrak{h}$ with $u'_i(0)=e_i$. It follow that the differential of the function $(u_1,\dots,u_m):\mathbf{R}^m\to H\subset\mathrm{SL}_n(\mathbf{R})$ at zero is surjective. Hence its image contains a neighborhood of $I_n$ in $\mathrm{SL}_n(\mathbf{R})$. Thus $H$ is open in $\mathrm{SL}_n(\mathbf{R})$ and hence by connectedness of the latter, is equal to $\mathrm{SL}_n(\mathbf{R})$.

Now remove the assumption that $H$ is in $\mathrm{SL}_n$. If $H\cap\mathrm{SL}_n=\mathrm{SO}_n$, then $H$ is contained in the normalizer $\mathbf{R}^*\mathrm{O}_n(\mathbf{R})$ of $\mathrm{SO}_n$. Otherwise, we deduce from the previous case that $H\cap\mathrm{SL}_n$ equals $\mathrm{SL}_n$, that is to say, $H$ contains $\mathrm{SL}_n$. This finishes the proof.