No. Let $f_n=x^2/n$ for $n$ odd and $(x_1)^2/n$ for $n$ even. Then $x_n^*$ is an alternating sequence of $1$s and $0$s, which does not converge to anything. But $f_n$ converges pointwise to $f=0$.
We can modify this to make the convergence uniform, by using an absolute value instead of a square, or to make $f$ nonconstant, by adding $\max(|x-1/2|,1)$ to $f_n$.
If $f$ has only a single zerounique minimum, the statement is true. Let $a$ be the $\lim\inf$ of $x_n^*$ and $b$ be the $\lim\sup$. Let $y$ be the unique minimum of $f$. Assume $ay$, a< y$. Then $f_n(a)$ converges to $f(a)$, and $f_n(y)$ converges to $f(y)$, and since $f(a)>f(y)$, $f_n(a)\leq f_n(y)$ only finitely many times. But every time $x_n^*\leq a$ we have $f_n(a)\leq f_n(y)$ since $a$ is closer to the minimum $x_n^*$ then $y$. This Since that occurs infinitely many times, this is a contradiction.
Therefore $a\geq y$. Similarly $y\geq b$, and by properties of $\lim\sup$ and $\lim\inf$ we have $b\geq a$, so $a=b=y$ and the limit is $y$.
