Martin provided one answer in the comments. Here’s another way to show that $d$ is a metric. For each $A\subseteq\Bbb N$ let $\chi_A$ be the indicator (or characteristic) function of $A$. Then
$$d(A,B)=\sum_{n\in\Bbb N}2^{-n}|\chi_A(n)-\chi_B(n)|\;$$
for any $A,B\subseteq\Bbb N$. (There’s no need to restrict oneself to finite subsets.) Hence
$$\begin{align*}
d(A,B)&=\sum_{n\in\Bbb N}2^{-n}|\chi_A(n)-\chi_B(n)|\\
&\le\sum_{n\in\Bbb N}2^{-n}\Big(|\chi_A(n)-\chi_C(n)|+|\chi_C(n)-\chi_B(n)|\Big)\\
&=\sum_{n\in\Bbb N}2^{-n}|\chi_A(n)-\chi_C(n)|+\sum_{n\in\Bbb N}2^{-n}|\chi_C(n)-\chi_B(n)|\\
&=d(A,C)+d(C,B)\;.
\end{align*}$$
This is actually just a special case of the proof that if $\{\langle X_n,d_n\rangle:n\in\Bbb N\}$ is a family of metric spaces such that each $d_n$ is bounded by $1$, and $X=\prod_{n\in\Bbb N}X_n$, then the function
$$d:X\times X\to\Bbb R:\langle x,y\rangle\mapsto\sum_{n\in\Bbb N}2^{-n}d_n(x_n,y_n)$$
satisfies the triangle inequality. (It is of course a metric on $X$.) Here each of the spaces $X_n$ is a copy of $\{0,1\}$ with the metric inherited from $\Bbb R$, and the map $A\mapsto\chi_A$ is an isometry.
No, this does not form a metric in general. For a specific counterexample, consider the case $X = [0, 1]^2$, $Y = \{ (x, y) \in [0, 1]^2 \mid x \le y \}$, and $f : X \to Y$ is defined by $f(x, y) = (xy, y)$. Then for all points in $Y$ with $y \ne 0$, the inverse image of $(x, y)$ is a single point $\{ (x/y, y) \}$; whereas the inverse image of $(0, 0)$ is $[0, 1] \times \{ 0 \}$. Thus, for example, in the induced $d_1$ on $Y$, we have $d_1((0, 0.1), (0.1, 0.1)) = d((0, 0.1), (1, 0.1)) = 1$. On the other hand, $d_1((0, 0.1), (0, 0)) = 0.1$ and also $d_1((0, 0), (0.1, 0.1)) = 0.1$.
You could try to fix this by considering paths, e.g. define $d_1(x, y)$ to be the infimum of values of the form $d(x^*, x_1) + d(y_1, x_2) + \cdots + d(y_{n-1}, x_n) + d(y_n, y^*)$ where $f(x^*) = x$, $f(y^*) = y$, and $f(x_i) = f(y_i)$ for each $i$. Then it would be straightforward to show this does satisfy the triangle inequality. However, then the tricky part would be to show that this infimum is strictly positive for $x \ne y$. I'm not sure off the top of my head how (or even if) that could be proved, though it seems clear that compactness would have to come into play in any proof of such a fact.
Best Answer
It's quite intuitive to show this from a slighly higher view:
For any set $S$ and the space $C_c(S, \mathbb R)$ (arbitrary functions with finite support), $\|f\| = \sum_{x\in S} |f(x)|$ is the well-know $\mathcal l^1$ norm. In particular, $d(f, g) = \|f-g\|$ is a metric over $C_c(S, \mathbb R)$.
Now we can embed the space of finite sets of $S$ to $C_c(S, \mathbb R)$ by mapping each finite subset $A$ to its characteristic function $\mathbb 1_A$. Then $|A\Delta B|= \|\mathbb 1_A-\mathbb 1_B\|$, hence as the latter is a metric, so is the former.