Def 1. A function $f:X \rightarrow \mathbb{R}$ is upper semicontinuous if $\forall x\in X:\forall \{x_n\}\subseteq X: x_n \rightarrow x \ \text{implies} \ f(x)\geq \limsup_{n\rightarrow\infty} f(x_n) $.
Def 2. A function $f:X \rightarrow \mathbb{R}$ is upper semicontinuous if $\forall x\in X:\forall \{x_n\}\subseteq X: x_n \rightarrow x \ \text{implies} \ \forall \epsilon >0:\exists N\in \mathbb{N}:\forall n>N:f(x)> f(x_n)-\epsilon $.
To show that both definitions are equivalent I decided to show the following: $$\ f(x)\geq \limsup_{n\rightarrow\infty} f(x_n) \iff \forall \epsilon >0:\exists N\in \mathbb{N}:\forall n>N:f(x)> f(x_n)-\epsilon.$$
My attempt
Let $x \in X$ and let $\{x_n\}$ be a sequence in $X$. Then
\begin{align} &f(x)\geq \limsup_{n\rightarrow \infty}f(x_n)=\lim_{n\rightarrow \infty} (\sup_{k\geq n} f(x_k))\\ \iff &\exists N\in \mathbb{N}:\forall n>N:f(x)\geq \sup_{k\geq n} f(x_k). \end{align} Note that $\sup_{k\geq n} f(x_k)$ forms a decreasing sequence along $n$. Furthermore, $\forall \epsilon>0$ and $\forall n\in \mathbb{N}$, $\sup_{k\geq n} f(x_k)\geq f(x_n)>f(x_n)-\epsilon $ . Thus
\begin{align} &\exists N\in \mathbb{N}:\forall n>N:f(x)\geq \sup_{k\geq n} f(x_k)\\ \iff & \forall \epsilon>0:\exists N\in \mathbb{N}:\forall n>N:f(x)>f(x_n)-\epsilon. \end{align}
Remark. To see the $(\Leftarrow)$ in the last equivalence above, note that as $\epsilon>0$ is arbitrary, we have $\forall n>N:f(x)\geq f(x_n)$, which implies $f(x)\geq \sup_{k\geq N}f(x_k)$, which in turn implies $\forall n>N, f(x)\geq \sup_{k\geq n}f(x_k)$, since $\sup_{k\geq n}f(x_k)$ is decreasing in $n$.
Do you agree with that?
Thanks!