Is every closed connected proper subgroup of $SO(2n+1)$ isomorphic (as a Lie group)
to a subgroup of $SO(2n)$? The answer is yes for abelian closed connected subgroups. 



Do all closed connected subgroups of $SO(2n+1)$ embed into $SO(2n)$?
No for $2n+1\ge 7$: $\mathfrak{so}(3)\oplus\mathfrak{so}(2n-2)$ does not embed into $\mathfrak{gl}(2n)$. 


Indeed start with $2n+1\ge 9$. Consider a nontrivial irreducible (real) representation of this product, nontrivial on both factors, of dimension $d$. It is isotypic for $\mathfrak{so}(3)$, say of the form $\mathfrak{v}\_k^{\oplus m}$ with $\mathfrak{v}\_k$ $k$-dimensional. If $k$ is odd, $\mathfrak{v}\_k$ is absolutely irreducible, and hence the centralizer is isomorphic to $\mathfrak{gl}\_m$, and hence $m\ge 2n-2$, so $$d=km\ge 3m\ge 3(2n-2)=6n-6=(2n+1)+(4n-5)\ge 2n+1.$$ Otherwise $k$ is a multiple of 4 (so $k\ge 4$) and the complexification is a sum of $2m$ $(k/2)$-dimensional irreducible representations, and hence the complexification of the centralizer is $\mathfrak{gl}\_{2m}(\mathbf{C})$. So $2m\ge 2n-2$, so 
$$d=km\ge 4m\ge 4(n-1)=4n-1\ge 2n+1.$$


Such irreducible being discarded, one remains with sums of irreducibles factoring through one quotient, and the smallest faithful one is clearly $(2n+1)$-dimensional.


For $2n+1=7$, one has to show that $\mathfrak{so}(3)\oplus\mathfrak{so}(4)\simeq\mathfrak{so}(3)^3$ does not embed into $\mathfrak{gl}(6)$. This is seen by hand: start with an irreducible of dimension $\le 6$, and suppose that it's nontrivial on the first factor. Then on the first factor, it's either $\mathfrak{v}\_3$, $\mathfrak{v}\_5$, $\mathfrak{v}\_4$ or $\mathfrak{v}\_3^{\oplus 2}$. The centralizer is abelian in the first two cases, and is isomorphic to $\mathfrak{so}\_3\times\mathbf{R}$ and $\mathfrak{gl}\_2$ in the last two cases. This shows that for $\mathfrak{so}(4)$, the only irreducible of dimension $\le 6$ (actually $\le 7$) that is nontrivial on both factors is 4-dimensional, and that there is no irreducible of $\mathfrak{so}(3)^3$ of dimension $\le 6$ that is nontrivial on each factor. If we allow only irreps that are nontrivial on only one factor, one has to add up to 9 to get faithfulness. So one irrep should be nontrivial on 2 factors, hence 4-dimensional. The remaining 2 dimensions are not enough to add anything.


Remains $2n+1=5$. At the level of Lie algebras there's no obstruction. Indeed any subalgebra of the 10-dimensional $\mathfrak{so}(5)$ has codimension $\ge 4$, that is dimension $\le 6$. Hence, if nonabelian it's isomorphic to $\mathfrak{so}(4)$, or $\mathfrak{so}(3)+\mathbf{R}^{k}$ for some $k$. A centralizer argument shows that $k\le 1$, so this indeed embeds into $\mathfrak{so}(4)$. I haven't checked at the group level.


