This definition of the boundary of $S_yf$ that you use
$$
\exists R>0: f(x) \le y, \forall x \in S_yf \cap B_R(\bar x) \setminus \{\bar x\}
$$
looks strange to me. It says that locally near $\bar x$ all $x$ from $S_yf$ should satisfy $f(x)\le y$, but this is true for all $x\in S_yf$ globally (by definition of the sublevel set), and it has nothing to do with the boundary.
Since you prefer to work with the finite values of $f$ we take $X=\operatorname{dom}f$. If $\operatorname{dom}f$ is a subset of a larger topological space then it becomes a topological space itself when equipped with the induced topology. Assuming $X$ to be Hausdorff we prove that the following statements are equivalent:
- $f\colon X\to\Bbb R$ is lower semicontinuous,
- $\operatorname{epi}f$ is closed in $X\times\Bbb R$.
Proof:
$\fbox{$1\Rightarrow 2$}$
Let $(x_\alpha,y_\alpha)_{\alpha\in A}$ be a net in $\operatorname{epi}f$ that converges to $(\bar x,\bar y)\in X\times\Bbb R$. We need to prove that $(\bar x,\bar y)\in\operatorname{epi}f$. But it is easily true because
$$
f(\bar x)\le\liminf f(x_\alpha)\le\liminf y_\alpha=\bar y.
$$
The first inequality is due to $f$ being lower semicontinuous, the second one comes from $(x_\alpha,y_\alpha)\in\operatorname{epi}f$.
$\fbox{$2\Rightarrow 1$}$
The set $\{(x,y)\colon x\in X\}\subset X\times\Bbb R$ is closed for each $y\in\Bbb R$. Hence, since $\operatorname{epi}f$ is closed, the intersection
$$
\operatorname{epi}f\cap\{(x,y)\colon x\in X\}=\{(x,y)\colon x\in X,f(x)\le y\}
$$
is closed too for each $y\in\Bbb R$. It follows that for each $y\in\Bbb R$ the projection on $X$
$$
S_y=\{x\colon x\in X,f(x)\le y\}
$$
is closed.
Now take a net $(x_\alpha)_{\alpha\in A}$ in $X$ that converges to $\bar x\in X$ and denote $\bar y=\liminf f(x_\alpha)$. This limit cannot be $-\infty$, because it would mean that $\bar x\in\operatorname{closure}(S_y)=S_y$ for every $y\in\Bbb R$, which is impossible (as $f(\bar x)\in\Bbb R$). Hence, $\bar y>-\infty$ and $\bar x\in\operatorname{closure}(S_{\bar y+\epsilon})=S_{\bar y+\epsilon}$ for all $\epsilon>0$. Taking $\epsilon\to 0$ gives that $f(\bar x)\le \bar y$. Hence, the function is lower semicontinuous.
The question has been updated in a comment.
False even on the real line. Can you think of a bounded non-empty set $M_0$ such that $I_{M_0}$ is not l.s.c.? Hint: take $M_0=(0,1]$.
Answer to the edited question: again consider the real line. if $M_0$ is closed then $I_{M_0}$ is upper semicontinuous, not lower semicontinuous. For example take $M_0=[0,1]$ and consider the sequence $x_n =1+\frac 1 n$. This sequence tends to $x=1$, $I_{M_0}(x_n)=0$ for all $n$, $I_{M_0}(x)=1$. Hence $I_{M_0}$ is not lower semicontinuous.
Best Answer
You are right. Lower semicontinuity is equivalent to: $$ \forall x \ \ \ \{ y| \delta_S(y)\le x \} $$is closed.
Let us prove this from OP's definition:
$\delta_S$ is lsc iff $\forall x\ \liminf_x \delta_S = \delta_S(x)$
Here this set is $\emptyset$ or $S$, so lower semicontinuity is equivalent to: $S$ is closed.