Open and closed set on a given metric I'm stuck on this one, could you please give me a tip or two:

Let $d$ be a metric in $X$ such that $d : X^2 \rightarrow \mathbb{R} : d(x, y) = d_{1}(x,y) + d_{2}(x,y)$ for $x,y \in X$. Let $d_{1}$ be a discrete metric. Show that in the metric space $(X,d)$ for any $x\in X$ the open metric ball $K(x, \frac{1}{2})$ of radius $\frac{1}{2}$ and centered in $x$ is a one-element set $\left\{x\right\}$. Also prove that any $A \subset X$ is both open and closed in the meaning of metric $d$.

Thanks in advance!
 A: The latter is simple enough, given the former, since if $\{x\}$ is open for all $x\in X$, then every subset of $X$ is open (as a union of singletons). It then follows that every subset of $X$ is closed. (Why?)
As for the former, take $x\in X$, and suppose that $y\in X$ such that $d(x,y)<\frac12.$ Since $d=d_1+d_2,$ then $d_1(x,y)<\frac12$--supposing that $d_2$ is a (pseudo)metric, anyway, so that $d_2(x,y)\ge0.$ By definition of the standard discrete metric, though, $$d(x,y)=\begin{cases}0 & x=y\\1 & x\ne y.\end{cases}$$ What can we conclude?

More generally, suppose that $d_1,d_2$ are any pseudometrics on $X$, and put $d=d_1+d_2$. It can readily be seen that $d$ is a pseudometric on $X$ as well. Now, for $x,y\in X$, we have $d_1(x,y)\le d_1(x,y)+d_2(x,y)=d(x,y)$ and likewise $d_2(x,y)\le d(x,y).$ Since $d$-distances are not smaller than $d_1$-distances or $d_2$-distances, then $d$-balls are not larger than $d_1$-balls or $d_2$-balls with the same center and radius.
In fact, we say even more than that: for any $x\in X$ and any $r>0,$ we find that the open $d$-ball of radius $r$ centered at $x$ is contained in the corresponding open $d_1$- and $d_2$-balls. Indeed, if $d(x,y)<r,$ then $d_1(x,y)\le d(x,y)<r,$ so the open $d$-ball of radius $r$ centered at $x$ is contained in the open $d_1$-ball of radius $r$ centered at $x$ (similar for $d_2$).
For example, define $d_1,d_2:\Bbb R^2\to\Bbb R$ by $d_1(v,w)=|v_1-w_1|$ and $d_2(v,w)=|v_2-w_2|$, where $v=\langle v_1,v_2\rangle$ and $w=\langle w_1,w_2\rangle$. Visually, $d_1$ is the pseudometric on $\Bbb R^2$ that only thinks about how much two points in the plane differ horizontally; $d_2$, only vertically. Neither of these is a metric--$d_1$ notices no difference between any two points on a vertical line, and $d_2$ notices no difference between any two points on a horizontal line. However, if we put $d=d_1+d_2,$ then $d$ is a metric sometimes known as the "taxicab metric."
The open $d_1$-ball of radius $1$ about the origin (for example) is the region of the plane lying strictly between the lines $x=-1$ and $x=1$. The open $d_2$-ball of radius $1$ about the origin is the region lying between $y=-1$ and $y=1$. The intersection of these two "balls" is the open square with vertices $\langle 1,\pm 1\rangle$ and $\langle-1,\pm 1\rangle$ (which, incidentally, is precisely the open ball of radius $1$ about the origin using the "chess king metric"), while the open $d$-ball about the origin of radius $1$ is the open square (or diamond, if you want to look at it that way) with vertices $\langle 0,\pm 1\rangle$ and $\langle\pm 1,0\rangle$ (a proper subset of the intersection of the corresponding $d_1$- and $d_2$-balls). In this case, we see that adding the two pseudometrics together actually loses us a bunch of points from corresponding balls of a given radius, since $d$ thinks that $d_1$- and $d_2$-balls are unbounded!

For a related example (though not precisely the same thing), suppose you're given a metric $d_1$ on $X,$ and a positive constant $c>1.$ Let $d=c\cdot d_1$. Then $d$ is a metric on $X,$ and the open $d$-ball of radius $r$ about $x$ is the open $d_1$-ball of radius $\frac{r}{c}$ (which is less than $r$) about $x$. Intuitively (from the "point of view" of $d_1$), $d$ has poor depth perception, and always sees objects as being farther apart than they "really" are but at least $d$ describes balls in the "right" way, so when $d$ is asked to describe a ball of radius $r$, he will describe a ball, but because $d$ thinks of the "actual" distance $r$ as $cr$, then the ball $d$ describes will be of "actual" radius $\frac rc.$
