In the case of (d) $\mathbb{N}$ is connected, since suppose we had a separation into two nonempty open disjoint subsets: $\mathbb{N}=U \cup V$. We know $U= \mathbb{N}-\{n_1,...,n_k\}$ and that $V=\mathbb{N}-\{m_1,...,m_l\}$. Now $U$ and $V$ can't be disjoint, as we can pick some natural number $n \notin \{n_1,...,n_k,m_1,...m_l\}$ and it is in both $U$ and $V$.
So I guess he is talking about components of subspace $U$.
That's right, Munkres refers to the connected components of the subspace $U$. Being connected is an intrinsic property of a topological space, it doesn't depend on any ambient space. So the components of $U$ are connected subsets of every space they are embedded in, in particular they are connected subsets of $X$, and therefore contained in components of $X$.
Note that since $U$ is open in $X$, for a subset of $U$ the conditions of being open in $X$ and being open in $U$ coincide. Since an open subspace of a locally connected space is again locally connected, one direction of the theorem can be formulated
The components of a locally connected space are open.
This follows almost immediately from the definition of local connectedness.
The other direction - if $X$ is a space such that the components of every open subspace $U$ of $X$ are open, then $X$ is locally connected - isn't hard either. Fix $x\in X$ and a neighbourhood $W$ of $x$. Since the components of $\overset{\Large\circ}{W}$ are open in $X$, the component of $\overset{\Large\circ}{W}$ that contains $x$ is a neighbourhood of $x$, and as a component, it is connected.
I have one more question: So how can we determine a component of $U$ in general. Is it equal to intersection of a component of $X$ and $U$?
In general, the intersection of a component of $X$ with a subspace $U$ need not be connected.
For a (very) simple example, consider the space $X = \{ z \in \mathbb{C} : \lvert z\rvert \notin \mathbb{N}\}$ and its open subspace $U = \bigl\{ z \in X : \lvert \operatorname{Re} z\rvert < \frac{1}{2}\bigr\}$. The components of $X$ are the annuli $A_n = \{ z : n < \lvert z\rvert < n+1\}$, and the intersection $A_n \cap U$ consists of two components for $n \geqslant 1$.
But don't be led astray, the general situation can be much more complicated. The intersection of a component of $X$ with an open subspace $U$ can have infinitely many components. As a more complicated but still simple example, let $X = \bigl\{ \bigl(t,\sin \frac{1}{t}\bigr) : t \in (0,+\infty)\bigr\} \cup \{0\}\times [-1,1]$ (the topologist's sine curve). Then $X$ is connected (but not path-connected), however the open subset $U = \{ (x,y) \in X : x^2 + y^2 < 1/4\}$ has infinitely many components. [Note: $X$ is not locally connected, the component $\{0\}\times (-1/2,1/2)$ of $U$ is not open in $U$ (in $X$).]
In general, the components of $U$ are subsets of the intersection of a component of $X$ with $U$. Occasionally, such an intersection will be connected, and hence a component of $U$, but typically it comprises many components of $U$. Still, knowing the components of $X$ can help identifying the components of $U$, since it can be easier to analyse the intersection of such a component with $U$ than to analyse the whole subspace $U$ (if the components of $X$ have a simple structure but $U$ is complicated).
Best Answer
A subsets $S$ of a topological space $X$ is a connected component of $X$ if and only if it is connected and it is a maximal element of the family of connected susets of $X$, ordered by inclusion $\subseteq$. Theis means that $S$ is connected, and that $S'$ is a connected subset such that $S'\supseteq S$ if and only if $S'=S$. If $X$ is a connected topological space, then the one and only connected component of $X$ is $X$ itself. In your example, neither of those sets $S_i$ is a connected component because neither of them sets is maximal by inclusion among the connected subsets of $S$.