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.
I think (1) is correct, but I wouldn't bother with the infima: let $a,b\in A$. Then
$$|d(x,a)-d(y,b)|\leq |d(x,a)-d(a,y)|+|d(a,y)-d(y,b)|\leq d(x,y)+d(a,b)$$
where we used the triangle inequality. Take infimum on both sides as $a\in A$, while $b$ is (for now) fixed, so $|f(x)-d(y,b)|\leq d(x,y)+0=d(x,y)$. Now this is true for all $b\in A$, so take infimum over $b$ and conclude that $|f(x)-f(y)|\leq d(x,y)$, I think this is more neat.
For (2) this is a standard topological fact: If $f:X\to Y$ is a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism. Indeed, all we need to do is show that $f^{-1}$ is continuous: if $E\subset X$ is a closed subset, we want to show that the inverse image of $E$ through $f^{-1}$ is closed in $Y$, which is the same as $f(E)$ is closed in $Y$. Now $X$ is compact and $E$ is a closed subset, so $E$ is also compact. Now $f$ is continuous, so $f(E)$ is a compact subset of $Y$. Now $Y$ is Hausdorff, so all compact sets are closed, thus $f(E)$ is closed in $Y$. Sometimes it helps viewing things in a more abstract frame; for example the metrics here are not essentially needed, so the extra information causes extra confusion.
For (3) you have a correct guess: Indeed, every element of the Cantor set $C$ can be written uniquely as $\sum_{k=1}^\infty\frac{a_k}{3^k}$, where $a_k$ is either $0$ or $2$. But it is important to note that the converse is also true: Any sum $\sum_{k=1}^\infty\frac{a_k}{3^k}$ where $a_k\in\{0,2\}$ is an element of the cantor set. Now define $f:C\to P$ as
$$f\big(\sum_{k=1}^\infty\frac{a_k}{3^k}\big)=(\frac{a_k}{2^{k+1}})_{k=1}^\infty$$
$f$ is well-defined: since $a_k$ is either $0$ or $2$, we have that $a_k/2^{k+1}$ is either $0$ or $1/2^k$, so each $a_k/2^{k+1}$ does indeed belong to $\{0,\frac{1}{2^k}\}$, so $f$ does indeed take values in $\prod_{k=1}^\infty\{0,\frac{1}{2^k}\}$.
$f$ is one to one: this is obvious.
$f$ is surjective: indeed, let $x=(x_k)_{k=1}^\infty\in P$, so each $x_k$ is either $0$ or $1/2^k$. Set $a_k=2^{k+1}\cdot x_k$. Then $a_k$ is either $0$ or $2$, so if $t=\sum_{k=1}^\infty\frac{a_k}{3^k}\in C$ then $f(t)=x$.
$f$ is continuous: Let $\varepsilon>0$ and fix a point $x\in C$, say $x=\sum_{k=1}^\infty\frac{a_k}{3^k}$, where $a_k\in\{0,2\}$.
Let $y\in C$ be another element, say $y=\sum_{k=1}^\infty\frac{b_k}{3^k}$ where $b_k\in\{0,2\}$. Note that
$$d_P(f(x),f(y))=\|\{\frac{a_k}{2^{k+1}}\}_{k=1}^\infty-\{\frac{b_k}{2^{k+1}}\}_{k=1}^\infty\|_{\ell^1}=\|\{\frac{a_k-b_k}{2^{k+1}}\}_{k=1}^\infty\|_{\ell^1}=\sum_{k=1}^\infty\frac{|a_k-b_k|}{2^{k+1}} $$
Claim: Let $m\geq1$ be an integer. Then if $|x-y|<\frac{1}{3^m}$, then $a_k=b_k$ for all $k=1,\dots,m$.
Proof of the claim: I will only describe the idea here. Note that the construction of the Cantor set is done in steps: in each step we divide the segments in 3 pieces and throw out the middle part. In the expression $x=\sum_{k=0}^\infty\frac{a_k}{3^k}$ with $a_k=0$ or $2$, we have the following interpretation: in the first step of the construction, we divide $[0,1]$ in three segments and we only keep $[0,1/3]$ and $[2/3,1]$. If $a_1=0$, then $x$ lies in $[0,1/3]$. If $a_1=2$, then $x$ lies in $[2/3,1]$. So if $|x-y|<1/3$ where $y\in C,y=\sum_{k=1}^\infty\frac{b_k}{3^k}$ with $b_k=0$ or $2$, then $b_1=a_1$: otherwise we would have $x\in[0,1/3]$ and $y\in[2/3,1]$ or $x\in [2/3,1]$ and $y\in[0,1/3]$, which contradicts the fact that $|x-y|<1/3$.
Let's assume that $a_1=0$, so $x\in[0,1/3]$. Now again, in the second step of the construction, we cut each of $[0,1/3]$ and $[2/3,1]$ in three pieces each and throw out the middle third of each. Let's look at $a_2$. If $a_2=0$, then $x$ lies in the left segment of $[0,1/3]$ that is left after the second step, i.e. $x\in[0,1/3^2]$. If $a_2=2$, then $x$ lies in the right segment of $[0,1/3]$ that is left after the second step, i.e. $x\in[2/3^2,1/3]$. Now if $|x-y|<1/3^2$ then we also have $|x-y|<1/3$, so $y$ lies in $[0,1/3]$ as we explained for the first step, i.e. $b_1=0$. Now if $b_2\neq a_2$, we have that $x,y$ are in different thirds of $[0,1/3]$ after the application of the second step of the construction, so $|x-y|\geq 1/3^2$, a contradiction. I believe that you get the idea.
Now let $m$ be so large that $\sum_{k={m+1}}^\infty\frac{1}{2^k}<\varepsilon$. Taking $\delta=\frac{1}{3^m}$; from the claim we have that if $y\in C$ satisfies $|x-y|<\delta$, then $a_k=b_k$ for all $k=1,\dots,m$, so
$$d_P(f(x),f(y))=\sum_{k=m+1}^\infty\frac{|a_k-b_k|}{2^{k+1}}\leq\sum_{k=m+1}^\infty\frac{2}{2^{k+1}}<\varepsilon$$
and continuity follows. Observe that $\delta$ is independent of $x$, so $f$ is uniformly continuous.
Is this good news? No, this is expected, since $f$ is continuous on a compact space, thus it must be uniformly continuous. Also, by (2) this gives us a homeomorphism of the cantor set and $P$.
Best Answer
a) Your counterexample is a good idea, but is too general. Being closed and open are not mutually disjoint properties, even for nonempty proper subsets. For instance, in $(0,1)\cup (2,3)$ with the subspace metric, $(0,1)$ is open and closed. In many examples, this will not work. For instance, take the specific inclusion $(0,1)\subseteq \mathbb R$. This is not a closed map as $(0,1)$ is closed on itself but not in $\mathbb R$
b) This is actually false. Consider $X=Y=\mathbb R$. Let $A = \{(x,y)\in \mathbb R^2 : xy=1\}$. This is the graph of $y=1/x$, so it is closed. However, $p_1[A] = \mathbb R-\{0\}$, which is not closed in $\mathbb R$.
c) Let $A\subseteq f[X]$ be closed. Then $A= f[f^{-1}[A]]$. As $f$ is continuous, $f^{-1}[A]$ is closed in $X$. Hence, $g[A]= g[f[f^{-1}[A]]]=(g \circ f)[f^{-1}[A]]$ is closed, as $g \circ f$ is closed by assumption.
d) The same counterexample as in part a) works. A closed ball in $(0,1)$ is just a closed interval $[a,b]\subseteq (0,1)$, which is certainly still closed in $\mathbb R$. However, as discussed in a) this inclusion map is not closed.