Suppose $f_1,f_2,\ldots $ is a sequence of convex functions that converges to a continuous convex $f$. Let $x_1^*,x_2^*$ be their respective (not necessarily unique) minima, and let y be a minima of $f$ (once again need not be unique). Can we prove that there exists a version of $x_1^*,x_2^*,\ldots$ such that $x_n^*\rightarrow y$ ?
No; here's a counterexample: let $f = 0$ and consider the minimizer $y = 0.$ Then you can construct convex functions which converge to $0$ pointwise but whose minima are always moving away from $y =0,$ e.g. $f_n(x) = (x - n)^2/n^n.$
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 a unique 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 $a< y$. 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 $y$. 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 $y$.