Q:

Can open sets be an open cover, for itself?

I have Baby Rudin's book with me and it clearly defines a cover to be open. In a followup, it defines a set $K$ to be compact if every open cover of $K$ contains a finite subcover.
And the rest I quote

More explicitly, the requirement is that if $\{ G_{\alpha}\}$ is an open cover of $K$, then there are finitely many indices $\alpha_1, \dots, \alpha_n$ such that $$K \subset G_{\alpha_1} \cup \dots\cup G_{\alpha_n}$$

From the notation, it would seem to suggest that we can't have $K$ to be an improper subset of the covers. Now in another book (which I referenced another user whom referenced the book) by Richardson (I don't know the full name sorry), it's definition of a cover allows a set to cover itself.
So what subtlety could I have possibly overlooked? Sorry if this mysterious reference is vague, but I couldn't get enough information on the book. 

A:

$K$ can be an improper subset in the definition. Some authors use $\subseteq$ and $\subset$ for subset and proper subset respectively, while others use $\subset$ and $\subsetneq$ respectively. Rudin uses the latter.
An open set obviously forms an open cover of itself, but a set is compact if every open cover has a finite subcover. You need to understand the notation for subset and definition of compact to avoid confusion.

A:

Note that a set can cover itself, but the requirement for a set to be compact states that every possible open cover has a finite subcover. As an example take $I=[0,1]\subseteq\mathbb{R}$ endowed with the subspace topology. Then every open cover $\{G_\alpha\}_{\alpha\in A}$ of $I$ will necessarily properly contain $I$ as $0$ and $1$ must be covered by some open intervals, say $(-\delta,\delta)$ and $(1-\varepsilon,1+\varepsilon)$ for $\delta,\varepsilon>0$. This says that $I\subsetneq \bigcup_{\alpha\in A}G_\alpha$ for every open cover, but since we're in a metric space and $[0,1]$ is closed and bounded, it is compact. Thus, there are a finite number of indices, say $\alpha_1,\ldots,\alpha_n$ such that $$I\subsetneq G_{\alpha_1}\cup\cdots\cup G_{\alpha_n}.$$As an alternative, if we consider $X=[0,1]$ as a topological space in its own right, then it is compact and hence every open cover $\{U_\alpha\}_{\alpha\in A}$ of $X$ will be such that $$X=\bigcup_{\alpha\in A}U_\alpha.$$ This shows that, depending on the space and whether or not it is viewed as a subspace, an open cover may properly contain the space or may equal the space.

A:

$K$ can be an improper subset in the definition. An open set obviously forms an open cover of itself, and is obviously a finite cover; but a set $K$ is compact if [and only if] every open cover has a finite subcover (as you yourself wrote, as per Definition 2.32 in Baby Rudin).
You'll find soon enough (if you have not already encountered this) that the definition of compactness you discuss above is the "foundational" definition of compactness used in Rudin. You'll see that chapter 2 proceeds to show that there are ways (using Theorems introduced and proven) to establish that a set is compact if it is, for example, (i) a closed subset of a compact space, or (ii) if it is in $\mathbb{R}^n$ and is closed and bounded ... etc.

I think the main confusion here is with the notation Rudin uses to denote "is a subset of."
Usually, authors (like Richardson) use $\subseteq$ to denote subset or set inclusion in general (meaning "is a proper subset of or is equal to") and they use $\subset$ exclusively to denote "is a (strictly) proper subset of". 
Other authors, like Rudin, simply use $\subset$ to denote the inclusion relation: "is a subset of or is equal to". Some, but not Rudin in PMA, use the symbol "$\subsetneq$" to denote the exclusion of set equality (i.e., to denote a proper subset).  See the "List of Special Symbols" which immediately precedes the "Index" in Baby Rudin: Rudin makes no mention of, nor any use of, the symbol $\subseteq$ nor of the symbol $\subsetneq$ in his text. 
So to specify "$A$ is a proper subset of $B$", Rudin will specify in some sense, that he is referring to a "proper subset" or otherwise omit the prospect of set equality, e.g. "$A\subset B$ AND $A \neq B$"). That is, unless Rudin states otherwise, assume $A\subset B$ means $A\subseteq B$ when reading the text.

