Assume $\Sigma$ is a linear space consisting of real square matrix of order n , $U$ is a subspace of $\Sigma$ and all nonzero elements in U are invertible, let
$$ \varphi(n)=\max_{U\subset\Sigma}\dim U $$
then the problem is to prove $\varphi(n)=1$ when $n$ is odd.
My attempts,
proof by contradiction, if there is two linearly independent elements called $A$ and $B$, then $cA+B \in U$.Then,
$$ cA+B \in U \Leftrightarrow |cA+B | = |A||cI_n + A^{-1}B| \ne 0 \quad(\forall c\in \mathbb{R}) $$ But the order is odd ,so $A^{-1}B$ has real eigenvalue c such that $|A||cI_n + A^{-1}B| = 0$, it's a contridiction, so $\varphi(n)=1$.
And my question,
Is there any other method and what happens when n is an even number?
For your first question: one alternative is the following, which uses determinants but not eigenvalues.
Suppose for contradiction that there are two linearly independent elements $A_0,B_0$. Set $A = A_0$ or $A = -A_0$, and similarly set $B = B_0$ or $B = -B_0$ so that $\det(A) > 0$ and $\det(B) < 0$ (keeping in mind that $\det(-M) = -\det(M)$ when $M$ has an odd size).
Consider the function $f:t \to \Bbb R$ given by $f(t) = |(1 - t)A + tB|$. Note that $f$ is a continuous function with $f(0) = |A| > 0$ and $f(1) = |B| < 0$. It follows that there is some $t_* \in (0,1)$ such that $f(t_*) = 0$. Thus, $(1 - t_*)A + t_*B$ is a non-invertible element of $U$, contradicting our premise.
A similar argument can be applied to the function $f(t) = |tA + B|$ corresponding to your proof, without any restriction on our choice of non-zero elements $A,B$. Rather than using the intermediate value theorem, it suffices to note that $f(t)$ is a real polynomial with odd degree.
For your second question: one class of examples comes from the real division algebras. So, there is the subspace of $\mathcal M_2(\Bbb R)$ given by $$ U = \left\{\pmatrix{a&-b\\b & a} : a,b \in \Bbb R\right\} $$ corresponding to the representation of the complex numbers, and a subspace of $\mathcal M_4(\Bbb R)$ that similarly corresponds to the representation of the quaternions.
These subspaces both satisfy the stronger requirement of being closed under matrix multiplication; I suspect that other examples exist that satisfy your requirement for $U$ but are not closed under multiplication.