Reflection relating two subspaces

What you seek to prove is false if you require the subspace $H$ to be a hyperplane.

To see this, let $S_1,S_2$ two planes in $\mathbb{R}^4$ that only meet in the origin (e.g. $S_1=span\{e_1,e_2\},$ $S_2=span\{e_3,e_4\}$), then for every hyperplane $H\subseteq\mathbb{R}^4$ we have $L\subseteq H\cap S_2$ for some line $L$ in $\mathbb{R}^4$, which means that $L\subseteq R_HS_2$, so $R_HS_2\neq S_1$.

What is true is that for two equidimensional subspaces $S_1,S_2\subseteq\mathbb{R}^n$ there is a subspace $T$ such that $R_TS_2=S_1$.

Let $\{u_1,\dots,u_h\}$ an orthonormal basis of $S_1\cap S_2$, $\{u_1,\dots,u_h,v_1,\dots,v_k\}$ an orthonormal basis of $S_1$, and $\{u_1,\dots,u_h,w_1,\dots,w_k\}$ an orthonormal basis of $S_2$. We put $T=span\{v_1+w_1,\dots,v_k+w_k\}$.

It is easy to see that $R_TS_2=S_1$. In fact we have $R_T(u_i)=-u_i$ since $T\perp(S_1\cap S_2)$ and $$R_T(w_i)=\Pi_T(w_i)-\Pi_{T^\perp}(w_i)=\frac{w_i+v_i}{2}-\frac{w_i-v_i}{2}=v_i$$ where $\Pi_T$ and $ \Pi_{T^\perp}$ are the orthogonal projections on $T$ and $T^\perp$, respectively. This concludes the proof. EDIT: This paragraph is wrong, see comments.

It's clear that the subspace $T$ is very much not unique. The presented choice for $T$ gives a subspace of minimal dimension with the desired property. If you are interested in a subspace of maximal dimension, a possible choice is $T'=T\oplus(S_1\cap S_2)\oplus(S_1+S_2)^\perp$

Here is a proof of the claim made by zvbxrpl using Riemannian geometry; it is a bit over the top as there should be just a linear algebra argument, I just do not have to to work out an elementary proof:

Theorem. For every pair of $k$-dimensional linear subspaces $P, Q$ in $E^n$ there exists an isometric involution $s=s_R$ whose fixed point set is a $k$-dimensional subspace $R$ in $E^n$ such that $s(P)=Q, s(Q)=P$.

Proof. The space of $k$-dimensional linear subspaces in $E^n$ can be equipped with the structure of a smooth manifold, called the Grassmannian $Gr_k(E^n)$. The latter admits a Riemannian metric making it into a symmetric space: For every $P\in Gr_k(E^n)$ there exists an isometric involution $s_P$ fixing $P$ pointwise such that the derivative of $s_P$ at $P$ is $-1$. More specifically, $s_P$ is the unique isometric (with respect to the euclidean metric on $E^n$) linear involution of $E^n$ fixing $P$ pointwise and having $-1$-eigenspace equal to the orthogonal complement to $P$.

See for instance pages 1-5 of these notes by Eschenburg where he works out this in detail.

Now, given two elements $P, Q\in Gr_k(E^n)$, there is a (possibly nonunique) minimizing geodesic $c$ connecting $P$ to $Q$. Let $R$ be the midpoint of this geodesic. Then the involution $s_R$ will send $c$ to itself and reverse its orientation, thus sending $P$ to $Q$ and $Q$ to $P$. qed