You are correct that its boundary as a subset of $\mathbb R^3$ is itself. However, there is another meaning of the word "boundary" being used by all the people you hear talking. As a two-dimensional surface, every point on the sphere has a neighborhood that "looks like" an open disk. A boundary point of a surface would be a point where all the neighborhoods have an "edge".
It is true that you have to think of neighborhoods within the sphere itself. This isn't so far fetched from our everyday experience. After all, when you think about the neighborhood you live in, you might think of homes within a quarter-mile "horizontal" radius, but you're probably not thinking about the air a quarter mile above your head or the rock a quarter mile below ground level. The neighborhoods within the sphere itself are analogous to neighborhoods as you would view them on a map, not as balls protruding above and below the sphere's surface.
If $S$ is the sphere, one way to define the open sets of $S$ relative to $S$ is using the subspace topology (or the restricted metric) from $\mathbb R^3$. An open "ball" centered at a point $x\in S$ has the form $\{y\in S:|x-y|<r\}$. Notice the extra qualification, "$y\in S$".
For contrast, let $D=\{(x,y)\in\mathbb R^2:x^2+y^2\leq 1\}$ be the closed unit disk in the plane. Thinking about the neighborhoods within $D$, there are two different kinds of points. Those points $(x,y)$ such that $x^2+y^2<1$ have neighborhoods that look just like the ordinary open disks in $\mathbb R^2$. But for a point $(x,y)$ with $x^2+y^2=1$, there is an asymmetry where all of the neighborhoods have $(x,y)$ on the "edge." The unit circle is the boundary of the disk as a surface, which coincidentally is also the boundary as a subset of the plane.
For another example, the top half of the sphere, with the equator included, is a surface whose boundary (as a surface) is the equator, whereas the boundary as a subset of $\mathbb R^3$ would be the whole set.
When you hear people say that the sphere is closed in this context, they are probably referring to closed surfaces, which by definition are compact and without boundary, where boundary is an intrinsic property of the surface as indicated above. (The linked article might have other useful information for you.) Of course, it is also true that the sphere is closed as a subset of $\mathbb R^3$.
Your proof consists of some correct steps done in the wrong order, which makes it something other than a valid proof. It looks more like scratchwork done in preparation for a proof. I rewrite it below, with some of the more important additions in bold. I will also change $t$ to $y$ throughout; when you wrote "$y$" you probably meant the same thing as "$t$".
Suppose $D(x_0, r)$ is a closed ball. We show that $X\setminus D(x_0,r) $ is open. In other words, for every point $y\in X\setminus D(x_0,r)$ we need to find an open ball contained in $X \setminus D$ with center $y$.
Since $y \in X\setminus D(x_0,r)$, it follows that $d(y,x_0) > r$, so $d(y,x_0) - r > 0 $. Let $r_1 = d(y,x_0)-r$.
I claim that the open ball $B(y, r_1)$ is contained in $X\setminus D(x_0,r)$. To prove this, consider any $z \in B(y,r_1)$. Notice by the triangle inequality
$$ d(x_0,y) \leq d(x_0,z) + d(z,y) \implies d(z,x_0) \geq d(x_0,y) - d(z,y) > d(x_0,y) - r_1 =r.$$
This shows $z\in X\setminus D(x_0,r)$, which completes the proof.
Best Answer
Easiest way to prove this is to first prove that the function $d_{x_0}: X \to \mathbb{R}$, defined as $d_{x_0} = d(x, x_0)$, is continuous (this follows essentially from the triangle inequality for the metric $d$), and then to note that the preimage of a closed set under a continuous function is closed, a singleton set $\{r\}$ is closed in $\mathbb{R}$, and $S(x_0, r) = d_{x_0}^{-1}(\{r\})$.