I am trying to prove that f is sequentially lower semicontinuous at x if and only if $f()=\sup_{r>0}\inf_{y \in B(x,r)}f(y)$.
Following the proof of ($F$ is lower semicontinuous $\iff F(x)=\sup_{r>0}\inf_{y\in B(x,r)}F(y)$ for all $x\in X$), I can understand the $<=$ implication but I cannot prove the converse direction.
For general function it holds $f(x)\geq \sup_{r>0}\inf_{y \in B(x,r)}f(y)$. So I need to show that if f is lower semicontinuous, then it holds $f(x)\leq \sup_{r>0}\inf_{y \in B(x,r)}f(y).$
Since for $x_n\to x$ it holds $\liminf_{n\to\infty}f(x_n)\geq f(x)$, it would suffice to find a sequence $(x_n)_n$ for which $\liminf_{n\to\infty}f(x_n) = \sup_{r>0}\inf_{y \in B(x,r)}f(y)$.
If $\sup_{r>0}\inf_{y\in B(x,r)} f(y)=f(x)$, then $f$ is lower semicontinuous at $x$. Let $\lim_{n\rightarrow \infty} x_n=x$. We define $r_n:=\sup_{m\geq n} \vert x _n -x \vert$. Note that $\lim_{n\rightarrow \infty} r_n=0$. Thus, we have \begin{align*} \liminf_{n\rightarrow \infty} f(x_n) &=\lim_{n\rightarrow \infty} \inf_{m\geq n} f(x_m) \geq \lim_{n\rightarrow \infty} \inf_{y\in B(x,r_n)} f(y) \\ &=\lim_{r\rightarrow 0^+} \inf_{y\in B(x,r)} f(y) =\sup_{r>0} \inf_{y\in B(x,r)} f(y) =f(x). \end{align*}
Now let's show the other direction. Namely, let $f$ by lower semicontinuous at $x$, then $\sup_{r>0}\inf_{y\in B(x,r)} f(y)=f(x)$. We have $\sup_{r>0} \inf_{y\in B(x,r)} f(y)\leq f(x)$ (this holds for every function), thus, we are left to show that $\sup_{r>0} \inf_{y\in B(x,r)} f(y)\geq f(x)$.
As $r\mapsto \inf_{y\in B(x,r)} f(y)$ is a decreasing function, we get $$ \sup_{r>0} \inf_{y\in B(x,r)} f(y) = \lim_{n\rightarrow \infty} \inf_{y\in B(x,1/n)} f(y).$$ For every $n\in \mathbb{N}_{\geq 1}$ pick $x_n \in B(x,1/n)$ such that $$ \inf_{y\in B(x,1/n)} f(y) \leq f(x_n) \leq \inf_{y\in B(x,1/n)} f(y)+ 1/n. $$ Then we have by the squeeze theorem $$ \lim_{n\rightarrow \infty} f(x_n) = \lim_{n\rightarrow \infty} \inf_{y\in B(x,1/n)} f(y) = \sup_{r>0} \inf_{y\in B(x,r)} f(y). $$ However, as we assumed that $f$ is lower semicontinous at $x$ we get $$ f(x)\leq \liminf_{n\rightarrow \infty} f(x_n) = \lim_{n\rightarrow \infty} f(x_n) = \sup_{r>0} \inf_{y\in B(x,r)} f(y). $$