The set of symmetric matrices that have multiple eigenvalues has Lebesgue measure 0. If the probability measure you use on the space of matrices is absolutely continuous w.r.t. the Lebesgue measure, then the probability that a random matrix has multiple eigenvalues is zero. I assume that this is the case. Assume that the matrices are $d\times d$.

$\newcommand{\bP}{\mathbb{P}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\eO}{\mathscr{O}}$ Suppose that your random matrices are defined on the probability space $(\Omega,\eO,\bP)$. Let $N\in\eO$ such that $\bP[N]=0$ and for any $\omega\in\Omega\setminus N$ all the matrices $M_n(\omega)$ have simple eigenvalues and

$$\lim_{n\to\infty} M_n(\omega)=D. $$

Denote by $\lambda_1(\omega)$ the smallest eigenvalue of $M_n(\omega)$ and by $\phi_n(\omega)$ an eigenvector of Euclidean norm $1$ corresponding to this eigenvalue. (There are two choices for $\phi_n(\omega)$ since $\lambda_1(\omega)$ is simple.) Then for any $\phi\in\bR^d$ of Euclidean norm $1$ we have

$$ \lambda_1(\omega)=\langle\; M_n(\omega)\phi_n(\omega),\phi_n(\omega)\;\rangle\leq \langle\; M_n(\omega)\phi,\phi\;\rangle. $$

If we let $n\to \infty$ along a subsequence $n_k$ such that $\phi_{n_k}(\omega)$ converges to some $\phi_\infty(\omega)$ of norm $1$ we deduce

$$ \langle\; D\phi_\infty(\omega),\phi_\infty(\omega)\;\rangle \leq \langle\; D\phi,\phi\; \rangle,\;\;\forall \phi\in\bR^d,\;\;\Vert\phi\Vert=1. $$

This proves that any limit vector $\phi_\infty(\omega)$ of the sequence $(\; \phi_n(\omega)\;)$ is an eigenvector of norm $1$ of $D$ corresponding to the lowest eigenvalue. That is the best that you can hope.

randomnessof the matrices $M_n$ that I gather are finite dimensional. If you assume that the entries of $M_n$ are continuous random variables then the result is true because in this case, with probability one, the spectrum of $M_n$ is simple. $\endgroup$