Proof that Compact set is Closed and Bounded

I came across the following problem while reading Spivak's Calculus on Manifolds.

Prove that a compact subset of $\mathbb{R}^n$ is closed and bounded.

The definition of an open set in the book is a set $A$ in which for all $x \in \mathbb{A}$, there exists an open rectangle $U$ such that $x \in U \subset A$. The definition of a closed set is a set $A$ in which $\mathbb{R}^n - A$ is open. The definition of a compact subset of $\mathbb{R}^n$ is a set $A$ in which for all open covers of $A$ (that is, collections of open sets that cover $A$), there exists a finite subcover of $A$. The boundary of a set $A$ is defined as the set of all points $x\in A$ such that for all open rectangles $U$ that contain $x$, $U$ contains points inside $A$ and outside $A$.

I attempted proof by contradiction first, asking myself a characteristic of not closed sets. Then I realized that if a set $A$ is closed, then $A$ must contain the boundary. I came up with the following proof

If $B$ is an open set and $x \in \text{Boundary}(B)$, then for all open rectangles $C$ such that $x\in C$, $C$ contains points of both $B$ and $\mathbb{R}^n - B$. This implies that $x \not\in B$. If $A$ is closed, then $\mathbb{R}^n - A$ is open, so the boundary of $A$ is not in $\mathbb{R}^n - A$, implying that the boundary of $A$ must be in $A$.

Now using this, I came up with a proof for the problem.

Let $A$ be a compact set. First, we show that $A$ must be bounded. Suppose that $A$ is not bounded. Then any finite open cover will only cover a finite volume, so this cannot hold. Now suppose that $A$ is not closed. Then there exists some $x$ in the boundary of $A$ such that $x$ is not in $A$. Now consider an open rectangle $U_1$ that contains $A$. Because an open rectangle is open, we must have that there exists some open rectangle $U_2$ that contains $x$ and is contained in $U_1$. By induction, we can show that there is a chain of open rectangles such that, $$x \in \cdots \subset U_3 \subset U_2 \subset U_1.$$ Now let $V_k = U_k + \text{Boundary } U_k$, which is closed. Then the open cover $$\{U_1 - V_3, U_2 - V_4, U_3- V_5, \dots \}$$ has no finite subcover since taking any cover out will exclude some boundary of an open rectangle.

The above argument could be made more rigorous. However, is the reasoning completely valid? Also, what is a simpler solution to the problem?

I think that your argument is simple enough, though it has some holes. First, you claim that any finite open cover has finite volume. This may not be true if the open sets are unbounded. Instead, construct an open cover (Say, the rectangle of side lengths all 1 centered at $x$ for all $x\in A$) which contains only bounded open sets. Then there still must exist a finite subcover. Then this argument works (assuming you can prove that bounded open sets have finite volume, and that finite volume implies boundedness). Perhaps an easier way than volume is to go back to the definition of boundedness. If your finite subcover is $\{U_1,...,U_n\}$, and for $x,y\in U_i$, $d(x,y)<1$, then for $x,y\in A$, the triangle inequality implies $d(x,y)\leq 2n$. This requires a little work, but it is not too hard.
For closed, this looks pretty good, but make sure that $V_{n+1}\subset U_n$. As stated right now, you could have $U_1=U_2=U_3=...$ so that $U_1-V_3=\emptyset$, so the cover you construct would not really be a cover.
Alternatively you could use a similar construct and collapse both cases ($A$ is open and $A$ is unbounded) into the same case. Your construct in the first case is basically that you construct a covering of "shells" around a limit point outside $A$, where each "shell" is essential to cover $A$.
For the case where $A$ is not closed we can build a sequence of annuli round a limit point outside $A$ in which case we find that a finite subcover of $A$ will have a inner radius and consequently leaving a bit around $x$ out.