Yes, your proofs are correct!
I think they could be more streamlined and readable, though. Due to the simplicity of the functions involved, it's easy to compute $v^{-1}(U)$ explicitly for $v = f, g, h$ and determine if the result is open in $\mathbb R$ rather than picking an arbitrary point $t \in v^{-1}(U)$ and then some $\delta> 0$ with $(t-\delta, t+\delta) \subseteq v^{-1}(U)$ to show that $v^{-1}(U)$ is open. In either case, it should first be justified continuity of $v$ is equivalent to "$v^{-1}(U)$ is open when $U$ is a basis element" (rather than an arbitrary open set) because we can always write $U$ as a union $U = \bigcup_\alpha U_\alpha$ of basis elements, and since inverse imeages preserve unions, $v^{-1}\left( \bigcup_\alpha U_\alpha \right) = \bigcup_\alpha v^{-1}(U_\alpha)$, the latter of which is open if $v^{-1}(U_\alpha)$ is open for basis elements $U_\alpha$.
By defining $v_n = \pi_n \circ v$, we can write $v \colon \mathbb R \to \mathbb R^\omega$ as $v = (v_1, v_2, \dots)$, where $v_n \colon \mathbb R \to \mathbb R$, in the sense that
$$
v(t) = \big( v_1(t), v_2(t), \dots \big).
$$
Substituting $v = f, g, h$,
\begin{align*}
f_n(t) = nt,\quad
g_n(t) = t,\quad
h_n(t) = \frac{t}{n}.
\end{align*}
We can consider the product and box topologies simultaneously by taking a general (nonempty) basis element
$$
U = (a_1, b_1) \times (a_2, b_2) \times \cdots,
$$
where $-\infty \leq a_n < b_n \leq \infty$ for all $n \in \mathbb N$. The product topology just has the stipulation that $(a_n, b_n) \neq (-\infty, \infty) = \mathbb R$ only for a finite subset of indices $n_1, \dots, n_k \in \mathbb N$.
To compute $v^{-1}(U)$, note that
$$
t \in v^{-1}(U) \iff v(t) \in U \iff \left[ v_n(t) \in (a_n, b_n) \quad\forall n \in \mathbb N \right].
$$
Hence
- $t \in f^{-1}(U) \iff nt \in (a_n, b_n) \quad \forall n \iff t \in \left( \tfrac{a_n}{n}, \tfrac{b_n}{n} \right) \quad \forall n \iff t \in \bigcap_{n=1}^\infty \left( \tfrac{a_n}{n}, \tfrac{b_n}{n} \right)$. The latter set is open in the product topology, being the intersection of only a finite number of open sets (since $(a_n, b_n) = \mathbb R$ for all but finitely many $n$), and hence $f$ is continuous in the product topology. However, that set is not always open in the box topology (take $a_n = -1$ and $b_n = 1$ for all $n$ to see $f^{-1}(U) = \{0\}$, which isn't open), and hence $f$ is not continuous in the box topology.
- $t \in g^{-1}(U) \iff t \in (a_n, b_n) \quad \forall n \iff t \in \bigcap_{n=1}^\infty \left( a_n, b_n \right)$. By the same reasoning as for $f$, $g$ is continuous in the product topology. In the box topology, taking $\left( a_n, b_n \right) = \left( -\tfrac{1}{n}, \tfrac{1}{n} \right)$ shows that $g$ is not continuous.
- $t \in h^{-1}(U) \iff \tfrac{t}{n} \in (a_n, b_n) \quad \forall n \iff t \in \left( n a_n, n b_n \right) \quad \forall n \iff t \in \bigcap_{n=1}^\infty \left( n a_n, n b_n \right)$. By the same reasoning, $h$ is continuous in the product topology. In the box topology, take $a_n = -\tfrac{1}{n^2}$ and $b_n = \tfrac{1}{n^2}$ to see that $h$ is not continuous.
For the uniform topology, I wouldn't change a thing!
It doesn't matter that $\lbrace u \in B_Y(y, \epsilon) : u \in \operatorname{Image}(f)\rbrace$ isn't open in $Y$; the inverse image of this set will be open.
You don't really need to split into two sets here either; just restrict the index set. If $\mathcal{V} \subseteq Y$ is open, then for each $y \in \mathcal{V}$, there exists an open ball $B(y, \varepsilon_y) \subseteq \mathcal{V}$. Then, for each $x \in f^{-1}(\mathcal{V})$, you can find a ball $B(x, \delta_x)$ such that
$$f(B(x, \delta_x)) \subseteq B(f(x), \varepsilon_{f(x)}) \iff B(x, \delta_x) \subseteq f^{-1}(B(f(x), \varepsilon_{f(x)})),$$
using the metric definition of continuity. Then
$$\bigcup_{x \in f^{-1}(\mathcal{V})} B(x, \delta_x) = f^{-1}(\mathcal{V}),$$
proving $f^{-1}(\mathcal{V})$ is open.
Best Answer
To have an answer: Yes, you are right, there's a $\lt \delta$ missing here.
In my seventh printing of the second edition of Munkres's book, this typo is fixed: