Since $m = \inf f$, there exists a sequence $x_n$ such that $f(x_n) \to m$. This is true just by the definition of $\inf$.
Since $x_n \in X$ and $X$ is compact, we have some $\hat{x} \in X$ and some subsequence such that $x_{n_k} \to \hat{x}$. This follows since $X$ is compact, and hence sequentially compact (equivalent in a metric space in any case).
Since $f$ is lsc., we have $\liminf_n f(x_{n_k}) \ge f(\hat{x})$. This is by definition of lsc.
Since $\liminf_n f(x_{n_k}) = \lim_n f(x_{n_k}) = m$, we have $m \ge f(\hat{x})$, and by definition of $m$, we have $m \le f(\hat{x})$, hence $\hat{x}$ is a minimizer.
Note: The above proof does not depend on $m$ being finite (the proof shows that $m$ is finite, since $m \ge f(\hat{x})$). However, it is straightforward to show that $m$ is finite directly.
Since $f$ ls lsc., at any $x$, for all $\epsilon>0$, there is a $\delta >0$ such that if $y \in B(x,\delta)$, then $f(y) > f(x)-\epsilon$.
So, choose $\epsilon=1$, then for each $x$ we have some $\delta_x >0$ such that if $y \in B(x,\delta_x)$, then $f(y) > f(x)-1$. The $B(x,\delta_x)$ form an open cover of $X$, hence by compactness, there is a finite subcover $B(x_k,\delta_{x_k})$. Hence $f(x) > \min(f(x_1),...,f(x_n))-1$, from which it follows that $m \ge \min(f(x_1),...,f(x_n))-1$.
A homeomorphism need not map open balls to open balls in a metric space.
But it does map open sets to open sets, as $f[O]=g^{-1}[O]$ where $g: Y \to X$ is the continuous inverse of $f$.
So if $N \in \mathcal{U}(x)$ it means that some metric ball $B_X(x,r) \subseteq N$ for some $r>0$ (I hope this agrees with your definition). $B_X(x,r)$ is open in $X$ (definition of the metric topology, or a lemma that's often proved right at the start of a topology course) so $f(x) \in f[B_X(x,r)] \subseteq f[N]$ and as that image of an open ball is open we have some ball in $Y$ with $f(x) \in B_Y(f(x),s) \subseteq f[B_X(x,r)] \subseteq f[N]$ so $f[N] \in \mathcal{N}(f(x))$ as required. Simply continuity of $f$ is enough to make $g[N]= f^{-1}[N]$ a neighbourhood of $x$ when $N$ is one for $f(x)$ so that the map $N \to f[N]$ is a bijection between $\mathcal{U}(x)$ and $\mathcal{U}(f(x))$.
b i) can be done more easily: if $A \subseteq X$ is closed, it's compact (being closed in the compact space $X$) and so $f[A]$ is compact by continuity of $f$, and thus closed in $Y$ (should be a well-known fact for you). So $f$ is a closed map.
And this implies quite easily that the inverse of a continuous bijection on a compact space is also continuous (continuity can also be seen as "the inverse image of a closed set is closed", just as we can for open sets). That part you already saw. But the sequences aren't really needed for b-i.
Best Answer
A proof not using contradiction:
Let $A_n := \{x \in X \mid f(x) < n\}.$ As $f$ is upper-semicontinuous, $A_n$ is an open set.
We have: $X = \cup_{n \in \mathbb N} A_n.$ $X$ is compact, therefore it has a finite subcover $F \subset \mathbb N$, i.e., $X = \cup_{n \in F} A_n$, which shows that $f$ is bounded from above by $M = \max_{n \in F} n.$