= Solution
For the reality of the <eigenvalues>, one may initially allow a nonzero complex <eigenvector> $\mathbf v$ with $M\mathbf v=\lambda\mathbf v$. Since $M$ is real symmetric, it is a <Hermitian matrix>, and
$$
\lambda=\frac{\mathbf v^\dagger M\mathbf v}{\mathbf v^\dagger\mathbf v}
$$
is real: the numerator equals its own complex conjugate and the denominator is positive. Once $\lambda$ is real, a nonzero real or imaginary part of $\mathbf v$ supplies a real eigenvector. We work with the real unit eigenvectors of the problem.
For two such <eigenvectors>, symmetry gives
$$
\lambda_a\,\mathbf e_a\cdot\mathbf e_b
=(M\mathbf e_a)\cdot\mathbf e_b
=\mathbf e_a\cdot(M\mathbf e_b)
=\lambda_b\,\mathbf e_a\cdot\mathbf e_b.
$$
Distinct <eigenvalues> therefore give zero <dot product>. Together with the prescribed unit lengths,
$$
\boxed{\mathbf e_a\cdot\mathbf e_b=\delta_{ab}.}
$$
These $n$ vectors form an <orthonormal basis> of $\mathbb R^n$.
Expand a real <unit vector> as $\mathbf x=\sum_ac_a\mathbf e_a$. Then $\sum_ac_a^2=1$, and its <Rayleigh quotient> is
$$
\mathbf x^TM\mathbf x=\sum_a\lambda_ac_a^2\le\lambda_1.
$$
Moreover,
$$
\lambda_1-\mathbf x^TM\mathbf x
=\sum_{a=2}^n(\lambda_1-\lambda_a)c_a^2.
$$
Every coefficient on the right is strictly positive, so equality forces $c_a=0$ for $a\ge2$. \b[Equality occurs precisely along the top eigendirection]:
$$
\boxed{\mathbf x^TM\mathbf x\le\lambda_1,\qquad
\mathbf x^TM\mathbf x=\lambda_1\iff\mathbf x=\pm\mathbf e_1.}
$$
To obtain a unit vector in $S$ using just $\mathbf e_1,\mathbf e_2$, let $u=(\mathbf e_1)_1$ and $v=(\mathbf e_2)_1$ denote their first coordinates. If $u^2+v^2>0$, take
$$
\alpha_1=\frac v{\sqrt{u^2+v^2}},\qquad
\alpha_2=-\frac u{\sqrt{u^2+v^2}}.
$$
Then $\alpha_1u+\alpha_2v=0$ and $\alpha_1^2+\alpha_2^2=1$. Hence $\alpha_1\mathbf e_1+\alpha_2\mathbf e_2$ has first coordinate zero and unit length. If $u=v=0$, simply take $\alpha_1=1,\alpha_2=0$. In either case the <Rayleigh quotient> of the constructed vector is
$$
\lambda_1\alpha_1^2+\lambda_2\alpha_2^2\ge\lambda_2.
$$
The continuous quadratic form attains its maximum on the <compact space> $S$, so
$$
\max_{\mathbf x\in S}\mathbf x^TM\mathbf x\ge\lambda_2.
$$
For $\mathbf x=(0,\mathbf y)^T$, deletion of the first row and column gives $\mathbf x^TM\mathbf x=\mathbf y^TA\mathbf y$, while $|\mathbf x|=|\mathbf y|$. The <spectral theorem for real symmetric matrices> applied to $A$ identifies its largest <eigenvalue> with this maximum:
$$
\mu=\max_{|\mathbf y|=1}\mathbf y^TA\mathbf y
=\max_{\mathbf x\in S}\mathbf x^TM\mathbf x.
$$
Combining the upper bound for every unit vector with the constructed lower bound proves
$$
\boxed{\lambda_1\ge\mu\ge\lambda_2.}
$$
This is <largest-eigenvalue interlacing for a principal submatrix>: restricting a quadratic form to a coordinate hyperplane cannot exceed its old maximum, but leaves some direction in the span of the top two eigenvectors.
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