If $K\subset Y$, does it necessarily mean that an open cover of $K$ exists in $Y$? The following doubt came to my mind when I was reading a proof that if $K\subset Y\subset X$ and $K$ is compact relative to $X$, then $K$ is compact relative to $Y$. 
$$$$The proof assumes that $K$ is compact relative to $X$ and supposes that $\{V_\alpha\}$ is a collection of sets open relative to $Y$ such that $K\subset \cup_\alpha V_\alpha$
Now this is where I got confused. I cannot understand why $Y$ will always have open subsets of itself such that their union covers $K$. I have also taken an example below to better illustrate the difficulty in understanding that I am facing.$$$$

If $K\subset Y$, does it necessarily mean that there exist open subsets $\{G_\alpha\}$ of $Y$ such that $K\subset \cup_\alpha G_\alpha$? In other words, does an open cover of $K$ always exist in $Y$ even if $K=Y$?

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I cannot see how it is always possible for a set $K\subset Y$ where $K=Y$ to have open subsets of $Y$ such that the union of these open subsets contains the set $K$ itself.$$$$
As a particular case, consider the set $K=\{z\in\mathbb{C}: |z|\le 1\}$ and let $Y=K$. Thus $K$ is a subset of $Y$. What are the $\textbf{open subsets }$ $\{G_\alpha\}$ of Y such that $K\subset \cup_\alpha G_\alpha$?
$$$$In essence my doubt arises because I cannot see how the union of $\textbf{open subsets}$ of a set can contain the set itself. $$$$I would be grateful if somebody would help me out. Many thanks in advance!
 A: There are two things I would like to bring up here. The first was mentioned by drhab in a comment: open covers with respect to $Y$ are made with sets that are open relative to $Y$. Hence, since $Y$ is always open in $Y$ by the definition of the subspace topology, $\{Y\}$ is always an open cover of $K$ under the conditions of the problem.
The larger point I would like to make is that the existence of open covers of $K$ is irrelevant to the compactness of $K$. A set is said to be compact if every open cover of it has a finite subcover. If there are no open covers of a set $S$, then all of the precisely $0$ open covers of $S$ would have finite subcovers and $S$ would be compact. This will never happen for the reasons above, but even if it did, it wouldn't present a problem to the proof.
A: Concerning your specific example, simply take $G=K=Y$. Then $G$ is an open subset of $Y$ and $K\subset G$.
Remember that, by definition, a subset $G$ of $Y$ is open when there is an open subset $O$ of $X$ such that $G=O\cap Y$. In the previous paragraph, just take $O=X=\mathbb C$ and you're done.
A: It seems that you are confusing "open subsets of $Y\subseteq X$" and "subsets of $Y$ that are open in $Y$".
If $U$ is an open subset of $X$ and this with $U\subseteq Y$ then you can call $U$ and open subset of $Y$.
But is this $U$ also an subset of $Y$ that is open in $Y$? Not necessarily.
Subsets of $Y$ that are open in $Y$ are by definition the sets that can be written as $U\cap Y$ where $U$ denotes an open subset of $X$.
These sets together form a topology on $Y$ which is called the subspace topology inherited from $X$.
Note that $Y$ itself is a set that is open in $Y$, because we can write $Y=X\cap Y$ and $X$ is an open subset of $X$.
Every set $K\subseteq Y$ is covered by the collection $\{Y\}$ and the elements of this cover (only one: $Y$ itself) are sets that are open in $Y$.
