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How to prove that the upper half plane, including the line $\{y=0\}$ with the subspace topology derived from the topology induced by euclidian open bols in $\mathbb{R}^2$ is separable and first-countable but not second-countable.

I proved that this space is separable and first-countable, but what to do to show that it is not second-countable? I suspect the problem arises in the line $\{y=0\}$. (I have followed a course in topology, some time ago.)

EDIT: because of the dicussion in the comment section I will give you the exact formulation of the problem: Endow the upper halfplane $M = \{(x, y) ∈ \mathbb{R}^2: y ≥ 0\}$ in $\mathbb{R}^2$ with a topological basis defined in the following way: we consider, for every point $(x, y) ∈ M$ with $y > 0$ and every $0 < \epsilon < y$, the neighbourhood $\{(z_1, z_2) ∈ M : ||(x − z_1, y − z_2)|| < \epsilon\}$; furthermore, for every $(x, 0) ∈ M$ and every $\epsilon > 0$, we consider the neighbourhood $\{x\} ∪ \{(z_1, z_2) ∈ M : ||(x − z_1, \epsilon − z_2)|| < \epsilon\}$. (In these definitions $|| · ||$ denotes the usual, Euclidean norm on $\mathbb{R}^2$.) We denote the resulting topological space by $(M, T )$.

Prove that $M$ is first-countable but not second-countable in the topology $\mathcal{T}$.

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    I think second countability is inherited for any subspace, whatever the topology. – 2017-02-22
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    @ Sassatelli Well, there are some more questions following these: I have to show the mentioned space has a subspace that is not separable and use these statements to conclude that it cannot be a metric space. – 2017-02-22
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    @user251257 thanks for mentioning, I meant $\mathbb{R}^2$. I have to conclude that the space is not metriziable – 2017-02-22

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The topology on $\operatorname{int}(M)$ is indeed the subspace topology from $\Bbb R^2$. But look carefully at the neighborhoods for points on the boundary. Those balls are not half-balls centered at $(x, 0)$, but instead are open balls tangent to the $x$-axis at $(x,0)$. This is why they have to add $(x, 0)$ specifically to these balls (your statement says "$\{x\}$, but that makes no sense as $x$ is just a real number; The point in $M$ is $(x,0)$). $(x,0)$ is on the boundary of the open ball, so is not itself inside. But since ball is intended to be a neighborhood of $(x,0)$, it has to be included.

To prove that it is not second-countable, note that the subspace toplogy the $x$-axis inherits from $M$ is discrete.