Here are a few hints and sketches of proofs. If the effective domain of $f$ is empty, then it is trivial, so assume it is nonempty.
For 1, if $x^*$ is not in the closure of the effective domain, then there exists an open ball $B$ containing $x^*$ such that $f(x)=\infty$ for all $x\in B$. Let $y$ be such that $F(x^*)=f(y) + \|x^*-y\|^2$. Note that $y\neq x^*$ as $y$ is in the effective domain of $f$. Then choose x'\in B closer to $y$ than $x^*$. Then
F(x') \leq f(y) + \|x'-y\|^2 < f(y) + \|x^*-y\|^2 = F(x^*)
which is a contradiction.
For 2, let $y$ be such that $F(x^*)=f(y)+\|x^*-y\|^2$. Since $F(x^*) \leq F(y) \leq f(y)$ it follows that $y=x^*$. So $F(x^*)=f(x^*)$. Since $\inf F \leq \inf f$, it follows that $x^* \in \arg\min\{f\}$.