You got confused about the definition of continuity.
If $f\colon Y\to X$ is continuous then the preimage of open subsets of $X$ is open in $Y$.
Since $X$ has the indiscrete topology, we only have two open subsets. Namely, $X$ and $\varnothing$.
The preimage of the empty set is of course empty, and therefore open in $Y$. If we look at $f^{-1}(X) = \{y\in Y\mid f(y)\in X\}=Y$, and of course that $Y$ is open in $Y$.
Thus, $f$ is continuous regardless to the topology given on $Y$ whenever $X$ is indiscrete.
Exercise: Suppose $f\colon X\to Y$ and $X$ has the discrete topology, prove that $f$ is continuous.
They are actually similar at a higher level. $\Bbb R$ has a subbase consisting of two types of sets: upper sets $U(a) =\{x \in \Bbb R: x > a\}$, and lower sets $L(a)= \{x \in \Bbb R: x < a\}$, where $a \in \Bbb R$. This holds in fact in all ordered topological spaces by the definition of the order topology.
Now $\{L(a),R(a): a \in \Bbb R\}$ generates the topology on $\Bbb R$ and in fact only the upper sets (plus $\{\emptyset, \Bbb R\}$) already form a topology on $\Bbb R$ (not a very nice one, only $T_0$ but not $T_1$ but it's sometimes used) and likewise for only the lower sets, and the usual topology is the supremum of these two weak topologies in the poset of all topologies on $\Bbb R$.
Now $f:X \to \Bbb R$ is upper semicontinuous (also as you state the definition) iff $f^{-1}[U(a)]$ is open in $X$ for all $a \in \Bbb R$. So it's continuity for the upper sets only. So quite literally semicontinuous: only the upper sets behave nicely under $f^{-1}$, hence upper semicontinuous. You could also say $f$ is continuous for the upper-topology on $\Bbb R$.
You can also check that lower semicontinuous functions are those that have $f^{-1}[L(a)]$ always open, and this explains their name. Also clear right away is that a function that is both upper and lower semicontinuous is continuous. Then openness of all inverse images of subbasic elements are open etc.
The same can be said for $\mathcal{P}(Y)$. This too has a standard topology that has a subbase that consists of two types of sets:
$[U]= \{F \in \mathcal{P}(Y): F \cap U \neq \emptyset\}$, where $U$ varies over all non-empty open subsets of $Y$ ("hit"-sets, because they have to "hit" $U$) and
$\langle U \rangle = \{F \in \mathcal{P}(Y): F \subseteq U\}$, where $U$ runs over the same sets ("miss"-sets as these are sets that "miss" (are disjoint from) $Y\setminus U$). All these types of sets together define a subbase for the Vietoris topology, which is normally restricted to the subspace of all closed (or compact even) non-empty sets of $Y$, to make it more nicely behaved, and is the major example of hit-and-miss topologies on power sets (here we have to hit open sets and miss closed sets, but other variants have been considered).
A set-values $F: X \to \mathcal{P}(Y)$ is upper semicontinuous if $F^{-1}[\langle V \rangle]=\{x \in X: F(x) \subseteq V\}$ is open in $X$ for all $V \subseteq Y$ open and non-empty. Again, continuity for half of the topology's subbase. It makes sense to call this the "upper" half as $\langle V \rangle$ are all sets upper bounded by $V$, as it were.
It's straightforward to check that it coincides with your definition (you take $x \in F^{-1}[\langle V \rangle]$ and the condition verifies it's an interior point of that set).
You can hopefully guess what a lower semi-continuous set-valued function must be, and check it against other statements of the definition.
You can also check that if $f:X \to \Bbb R$ is seen as a set valued function, e.g. by $F(x) = \{(-\infty,f(x)]\} \in \mathcal{P}(\Bbb R)$ then usc-ness of $F$ comes down to usc-ness of $f$ and vice versa (please check that, but it seems OK).
So actually, the definitions are similar: it's continuity wrt a half of a standard subbase of the codomain.
Best Answer
Consider the function $$f \colon \mathbb{R} \to \mathbb{R}, \, x \mapsto \begin{cases} 0, & x \in \mathbb{Q} \\ 1, & x \notin \mathbb{Q} \end{cases}$$ where $\mathbb{R}$ is equipped with the Euclidean topology. For any subset $U \subseteq \mathbb{R}$ the preimage $$f^{-1}(U) \in \{\emptyset, \mathbb{Q}, \mathbb{R} \setminus \mathbb{Q}, \mathbb{R}\}$$ then is b-open since $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$, so $f$ is b-continuous.
However, $f$ is not semi-continuous: The interval $U = (-\tfrac{1}{2}, \tfrac{1}{2}) \subseteq \mathbb{R}$ is open with $f(0) = 0 \in U$, but for any semi-open $V \subseteq \mathbb{R}$ with $f(V) \subseteq U$ we have $$V \subseteq \operatorname{cl}(\operatorname{Int}(V)) \subseteq \operatorname{cl}(\operatorname{Int}(f^{-1}(U))) = \operatorname{cl}(\operatorname{Int}(\mathbb{Q})) = \operatorname{cl}(\emptyset) = \emptyset$$ and thus $0 \notin V.$
Regarding your trivial attempt: Any function mapping to a target space equipped with the indiscrete topology is continuous and hence also b-continuous and semi-continuous.