I'm doing Problem III.3.9 from textbook Analysis I by Amann.
Could you please verify if my proof looks fine or contains logical gaps/errors? Any suggestion is greatly appreciated.
My attempt:
(a)
(i) $\implies$ (ii): Assume the contrary that there are $x_o \in X$ and $\varepsilon>0$ such that $$\forall U \in \mathcal{U}(x_0), \exists x \in U: f(x) \le f(x_0) - \varepsilon$$
By Axiom of Countable Choice, there is a sequence $(x_n)$ in $X$ such that $f(x_n) \le f(x_0) - \varepsilon$. Then $\varliminf f(x_n) \le f(x_0) - \varepsilon$ and consequently $\varliminf f(x_n) \le f(x_0) - \varepsilon < f(x_0)$, which contradicts the fact that $f$ is lower continuous.
(ii) $\implies$ (iii): For $x_o \in f^{-1} [(\alpha,\infty)]$, there is $U \in \mathcal{U}(x_0)$ such that $f(x) > f(x_0) - \varepsilon$ where $\varepsilon = f(x_0) -\alpha>0$ for all $x \in U$. Then $f(x) > \alpha$ for all $x \in U$ and hence $U \subseteq f^{-1} [(\alpha,\infty)]$. As such, $f^{-1} [(\alpha,\infty)]$ is open.
(iii) $\implies$ (iv): We have $X - f^{-1} [(-\infty,\alpha]] = f^{-1} [\mathbb R] - f^{-1} [(-\infty,\alpha]] = f^{-1} [\mathbb R - (-\infty,\alpha]] = f^{-1} [(\alpha,\infty)]$, which is open. As such, $f^{-1} [(-\infty,\alpha)]$ is closed.
(iv) $\implies$ (i): Assume the contrary that there is $\alpha \in \mathbb R$ such that $A=f^{-1}[(-\infty,\alpha]]$ is open. For every sequence $(x_n)$ in $A$ such that $x_n \to b \in X$, $f(b) \le \varliminf f(x_n) \le \alpha$ because $f(x_n) \le \alpha$. As such, $b \in A$ and hence $A$ is closed, which is a contradiction. As a result, $f^{-1}[(-\infty,\alpha]]$ is closed for all $\alpha \in \mathbb R$.
(b)
$\Longrightarrow$: If $f$ is continuous at $x$ than $f(x) = \lim f(x_n) = \varliminf f(x_n) = \varlimsup f(x_n)$ for all $(x_n) \to x$. As such, $f(x) \le \varliminf f(x_n)$ and $-f(x) \le - \varlimsup f(x_n)= \varliminf (-f(x_n))$. Hence $f$ is lower and upper continuous.
$\Longleftarrow$: If $f$ is lower and upper continuous, then $f(x) \le \varliminf f(x_n)$ and $-f(x) \le \varliminf (-f(x_n)) = -\varlimsup f(x_n)$ for all $(x_n) \to x$. Hence $\varlimsup f(x_n) \le f(x) \le \varliminf f(x_n)$ and thus $f(x) = \lim f(x_n)$. As such, $f$ is continuous at $x$.
(c)
We have $\chi_{A}:X \to \mathbb R$ where $\chi_{A}(x)=1$ if $x \in A$ and $0$ otherwise.
$\Longrightarrow$: Let $(x_n)$ be a sequence in $X$ such that $\lim x_n = x$. If $x \in A$ then there is a neighborhood $U$ of $x$ such that $U \subseteq A$. Then $\chi_{A}(x_n) = 1$ for all $x_n \in U$. Thus $\chi_{A}(x) =1 =\varliminf \chi_{A}(x_n)$. If $x \notin A$, then $\chi_{A}(x) =0 \le \varliminf \chi_{A}(x_n)$. As such, $\chi_{A}$ is lower continuous.
$\Longleftarrow$: If $\chi_{A}$ is lower continuous, then $f^{-1}[(0,\infty)] = A$ is open by (a-iii).
(d)
By definition of infimum, there is a sequence $(y_n)$ in $f[X]$ such that $y_n \to \inf f[X]$. By Axiom of Countable Choice, there is a sequence $(x_n)$ in $X$ such that $f(x_n) = y_n$ for all $n$. Because $X$ is compact, there is a subsequence $(x_{n_m})$ of $(x_n)$ such that $x_{n_m} \to x \in X$.
Because $f$ is lower continuous, $f(x) \le \varliminf f(x_n)$. On the other hand, $\varliminf f(x_n) \le \varliminf f(x_{n_m}) = \varliminf y_{n_m} = \lim y_{n_m} = \lim y_n = \inf f[X]$. As such, $\inf f[X] \le f(x) \le \inf f[X]$. Hence $\inf f[X] =f(x)$. As such, $f$ attains its minimum.