The path-components of $F$ are the singletons in $G$, as well as horizontal lines (more precisely positive open half-rays) in $H$. On the other hand, $F=G\cup H$ is connected (and hence is its own component, just one.)
To prove the statement about path-components, say there was a path from a point $h\in H$ to a point $g\in G$, say $h=(z,p)$ and $g=(0,q)$, where $z>0$, $p$ is irrational, $q$ is rational, and both $p,q\in[0,1]$. We may assume without loss of generality that $\gamma:[0,1]\to F$ is a path with $\gamma(0)=h$ and $\gamma(1)=g$. Let $L_p:=\{(x,p):x>0\}$ and define $T=\{t\ge0:\gamma([0,t])\subseteq L_p\}$ and $s=\sup T.$ Left to the reader (ask for details if needed): the set $T$ is both open and closed, and $0\in T$, hence $T=[0,1]$ and $s=1$, hence (by continuity) $\gamma(1)\in L_p$, contradicting $\gamma(1)=g$.
The above shows more generally that if $\gamma$ is any path containing at least one point in the set $L_p:=\{(x,p):x>0\}$ (for some fixed irrational $p\in[0,1]$) then $\gamma([0,1])\subset L_p$. It remains to show that if $q\in[0,1]$ is rational, then singleton $\{(0,q)\}$ is a path-component of $F$. The above argument shows that no path starting at $(0,q)$ could contain any point in $H$. But it is quite obvious that if the path has to stay in $G$ all the time, then it has to stay at $(0,q)$ all the time.
It remains to show that $F$ is connected. Suppose there was a non-empty closed-and-open set $U\subseteq F,$ we will show that $U=F.$ As before, if $p$ is any irrational in $[0,1]$, let $L_p:=\{(x,p):x>0\}$. Observe that for any irrational $p$ in $[0,1]$, if $\varnothing\not= L_p\cap U$ then $L_p\subset U$. We cannot have $U\subseteq G$ (since then the non-empty $U$ could not be open in $F$), hence it follows that $\varnothing\not= H\cap U$. Fix any $z>0$ and irrational $p$ in $[0,1]$ with $h:=(z,p)\in H\cap U.$ Let $r=\sup Y$ where $Y=\{y\ge p: y \in [0,1]\setminus\mathbb{Q},\ L_y\subset U\}.$ If $r$ is irrational, then using that $U$ is closed, we must have $r\in Y$, but then using that $U$ is open one easily shows that there must be $\varepsilon>0$ with $r+\varepsilon\in Y$, contradicting that $r=\sup Y.$ Hence $r$ is rational. We will show that $r=1.$ Suppose to the contrary that $p<r<1.$ Since $U$ is closed (and since $L_y\subset U$ for every irrational $y$ with $p\le y<r$) we must have $(0,r)\in U.$ But, since $U$ is open, some neighborhood $N$ of $(0,r)$ must be contained in $U$, and, using that $r<1$, it follows that $N$ must contain the points $(\delta, r+\delta)$ for all small enough $\delta>0$ (and with $\delta$ irrational).
Hence $U$ must contain $L_{r+\delta}$ for all small enough irrational $\delta>0$, contradicting the definition of $r=\sup Y.$ This proves $r=1$ and it follows that every point $(a,b)$ in $F$ with $a\ge0$ and $p\le b\le1$ must belong to $U.$ In a similar way one proves that every point $(a,b)$ in $F$ with $a\ge0$ and $p\ge b\ge0$ must belong to $U,$ hence $U=F.$ This completes the proof that $F$ is connected, a nice problem, thank you!
Edit 12 July, 2021.
Addressing the following comments.
Comment (to be clarified). Could you provide more details on why $T$ is both open and closed (by using the definitions and giving formal proofs, ideally)?
More details. So, $T=\{t\ge0:\gamma([0,t])\subseteq L_p\}$ (where, as above, $\gamma(0)=h=(z,p)$ and $L_p:=\{(x,p):x>0\}$). We first show that $T$ is closed.
Clearly $0\in T$ and if $t\in T$ and $0\le t'<t$ then also $t'\in T$. It follows that if $s=\sup T$, then $T$ is equal to either $[0,s)$ or $[0,s]$. (Note that if $s=0$ then clearly $s\in T=\{0\}$ which is closed, so we may assume below that $s>0.$) We will show that $T=[0,s]$ (which is clearly closed). We definitely have that $[0,s)\subseteq T$ (since if $0\le t'<s$ then there is some $t\in T$ with $t'<t<s$, hence $t'\in[0,t]\subseteq T$). By continuity, $\gamma(s)=\lim\limits_{t\nearrow s}\gamma(t).$ Let $\gamma(t)=(\xi(t),\eta(t))$, that is $\xi(t)$ and $\eta(t)$ denote the $x$- and the $y$-coordinates of $\gamma(t)$, respectively.
We have $\eta(s)=\lim\limits_{t\nearrow s}\eta(t)=\lim\limits_{t\nearrow s} p=p.$ That is, $\gamma(s)=(\xi(s),p)$ and it has to be the case that $\xi(s)>0$ and $\gamma(s)\in L_p.$ Hence $s\in T$ and $T=[0,s]$ which completes the proof that $T$ is closed. Next we prove that $T$ is open in the relative topology of $[0,1].$ If $s=1$ there is nothing to prove since $T=[0,1]$ which is open in $[0,1].$ So, assume it were the case that $0\le s<1.$ Since $\xi(s)>0$, by continuity there is some $\varepsilon>0$ such that $\xi(s+\delta)>0$
whenever $0\le\delta\le\varepsilon.$ We will show that $(s,s+\varepsilon]\subseteq T.$ Assume to the contrary that there is some $t'$ with $s<t'\le s+\varepsilon$ such that $\gamma(t')\not\in L_p.$ This means that if $p'=\eta(t')$ then $p'\not=p.$
But then the restriction $\gamma:[s,t']\to H$ would be a continuous path remaining in $H$ all the time and connecting a point in $L_p$ to a point in $L_{p'}$ which is a contradiction since $L_p$ and $L_{p'}$ are different components of $H.$ This completes the proof that $T$ is open (and, hence, that it is both closed and open).
Comment (to be clarified). Also, could you formalize the statement "but it is quite obvious that if the path has to stay in $G$ all the time, then it has to stay at $(0,q)$ all the time."?
Formalization (and justification).
By "the path has to stay in $G$ all the time" we mean that $\gamma(t)\in G$ for all $t\in[0,1]$ or in other words $\gamma([0,1])\subseteq G$, or we could also say $\gamma:[0,1]\to G$ (rather than the correct but not so specific $\gamma:[0,1]\to F$). Similarly, by
"It has to stay at $(0,q)$ all the time" we mean that it has to be the constant path $\gamma(t)=g=(0,q)$ for all $t\in[0,1].$
Indeed, if $\gamma:[0,1]\to G$ were not constant, then there is some $t'$ such that $\gamma(t')=g'=(0,q')\not=(0,q).$ But then the restriction $\gamma:[0,t']\to G$ would be a path in $G$ connecting $(0,q)$ to $(0,q')$ which is a contradiction (since the components, and hence the path-components of $G$ are singletons).
Comment (to be clarified). Also, could you further justify the statement "If $r$ is irrational, then using that $U$ is closed, we must have $r\in Y$, but then using that $U$ is open one easily shows that there must be $\varepsilon>0$ with $r+\varepsilon\in Y$, contradicting that $r=\sup Y.$"?
Further justification. By construction, $r\ge p.$ If $r$ were irrational and if $r\not\in Y,$ then $r>p$ (since $p\in Y$). Consider the point $(1,r)$ (where $1$ is picked somewhat arbitrary, any positive number in its place would do). Pick $y_n\in Y$ with $p<y_0<...<y_n<y_{n+1}<...<r$ and with $\lim\limits_{n\to\infty}y_n=r.$ Then $(1,y_n)\in U$ for all $n$, and the sequence of points $(1,y_n)$ converges to the point $(1,r)$, and since $U$ is closed it follows that $(1,r)\in U$. This in turn implies that $L_r\subseteq U$ and hence $r\in Y.$ Next we clarify that part of the statement that says: "but then using that $U$ is open one easily shows that there must be $\varepsilon>0$ with $r+\varepsilon\in Y.$" Indeed, since $U$ is open, there must be some open ball $B$ centered at $(1,r)$ such that $B\cap F\subseteq U$. Then $(1,r+\varepsilon)\in U$ whenever $r+\varepsilon$ is irrational and $\varepsilon$ is smaller than the radius of $B$. For such $\varepsilon$ we must have $r+\varepsilon\in Y.$
Best Answer
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 $V$.