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In the definition of the Moore plane X=L1∪L2, where L1 is the line y=0 and L2=X∖L1 , I have a problem. How can i show using Subspace Topology that L1 is discreet?

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The answer to this question depends on how you defined the topology.

However, if $p = (x,0) \in L_1$, then a fundamental system of open neighborhoods for $p$ are balls centered at $(x,\epsilon)$ with radius $\epsilon$ (union the tiny point $p$).

When you intersect this open neighborhood with $L_1$, you obtain the singleton $\{p\}$!

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    still don't understand, and can closed **L1** be discreet? – 2017-02-14
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    @Duboskydumont $\mathbb{Z}$ in the reals is also closed and discrete, as are all finite subsets of a $T_1$ space. So closed and discrete combined is not uncommon – 2017-02-15
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    @HennoBrandsma you mean this http://math.stackexchange.com/questions/1270822/why-is-the-set-of-integers-in-mathbbr-closed – 2017-02-15
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    What does this have to do with the problem? Discreet just means that it has the discreet topology, namely every point of $L_1$ is open in $L_1$, and this is exactly what the argument above shows. – 2017-02-15
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    @fulges that all interger number is closed that is for HennoBrandsma – 2017-02-17
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A basic neighbourhood of a point $(x,y)$ where $y > 0$ is a classical (standard metric) ball of the plane $B((x, y), r)$, where $r \le |y|$. Note that all these balls miss $L_1$, so $X \setminus L_1$ is open, and $L_1$ is closed.

A basic neighbourhood of a point of the form $(x,0)$ is of the form

$$N_r((x,0)) = \{(x,0)\} \cup B((x,r), r),r > 0 \text{ and note : } N_r((x,0) \cap L_1 = \{(x,0)\}$$

which makes it clear that $\{(x,0)\}$ is open in $L_1$ by the definition of the subspace topology. A subspace is discrete iff each of its points is an isolated point in the subspace topology. So $L_1$ is closed and discrete, while it's clear that both the balls $B((x, y),r)$ and $N_r((x,0))$ intersect $\{(x,y): x,y \in \mathbb{Q}, y > 0\}$, which is thus a countable dense set.

So Jones' lemma will imply that $X$ is not normal.

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    i Did not understand "Note that all these balls miss **L1** " can u explain me that part – 2017-02-15
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    @Duboskydumont You do see that $B((x,y), r) \cap L_1 =\emptyset$ if $r \le y$? This is a simple computation. – 2017-02-15
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    @HennoBrandsmayes i see, what u mean with that? – 2017-02-15
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    It means every point not in $L_1$ has a neighbourhood missing it. This is what implies that $L_1$ is closed – 2017-02-16
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    @HennoBrandsmayes thanks – 2017-02-17