Prove that it is first countable Hello I have problems with this exercise
Let $H:=\{(x,y)\in \mathbb{R}^2\;:\; y>0\}$ and $R=\{(x,y)\in  \mathbb{R}^2\;:\; y=0\}$. Notice that $\tau$ he topology of $H$ induced by $ \mathbb{R}^2$. Let the set  $X(S):=H\cup S$, where  $S\subset R$.
Define the topology $\tau^{\star}$ over the set  $X(S)$ as  the generated topology $\tau$ and the set $\mathcal B$, where $\mathcal{B}$ is formed by all the sets of form$\{x\}\cup B$, where $x\in S$ and $B\subset H$ is an open ball tangent to $R$ at the point $\{x\}$.
a) Let $S'\subset S$. Verify that $S'$ is a closed subset of $X(S)$.
b) Let $F'$ a closed subset of $\mathbb{R}^2$, verify that  $F:=F'\cap X(S)$  is a closed subset of $X(S)$
c) Prove that $X(S)$ is first countable and regular
d) Assume that $S$ is a countable set. Prove that $X(S)$  is a second countable space. Is  $X(S)$ metrizable?
My attempt
I made a drawing to understand better
a)    We must show $X(S)^{c}$ is open. Let $x \in X(S)^{c}$ . By definition this means that some open neighborhood $U_x$ of $x$ either does not meet $S$ or does not meet $S^c $. Since $U_x$ is an open neighborhood of each of its points, every point $u \in U_x$ is in $X(S)^{c}$ and hence $U_x \subset X(S)^{c}$ . Therefore $X(S)^{c}$ is open since it contains an open neighborhood of each of its points . It's right?
b) I don't know
c)  I know that:
A topological space satisfies the first countability axiom, or is first countable, if every point admits a countable local basis of neighbourhoods. but I don't know what local base I can find
d) How can I prove that it is compact?
Thanks
 A: As to a): You have to verify that $S'$ is closed, so let $(x,y) \in X(S)\setminus S'$. If $(x,y) \in H$ then $H$ itself is open in $X(S)$ and is disjoint from $S'$ (which lies inside $S$ so inside $R$). So the only interesting case is when the point is $(x,0)$ which is not in $S'$. But then any tangent open disk at $(x,0)$ together with $\{(x,0)\}$ is by definition a basic open subset of $X(S)$ that only intersects $R$ in $(x,0)$ so is disjoint from $S'$ too!
b) is quite trivial if you think about it. $F$ has two parts, one in $H$, one in $R$, and in $R$ we know from (a) that the part in $R$ is always closed while the part in $H$ is closed there too it has the subspace topology wrt the plane already).
c) is also trivial: in $H$ we have the standard metric local countable base and for $R$ we restrict ourselves similarly to tangent circles of radius $\frac1n$ only, for $n \in \Bbb N$. This obviously works as a local base. Regularity follows from simple reasoning and some case distinguishing (points in $R$ vs points in $H$ etc.)
d) $H$ is second countable already and we add the countably many countable local bases at $S$ to get a total countable base. So metrisability follows immediately from Urysohn and c). You cannot show that $X(S)$ is compact because it's not. $H$ is already unbounded, e.g. and $S$ is closed and discrete...
