# Space is both $T_1$-space and normal space is Hausdorff space

Every metric space which is both $$T_1$$-space and normal space is Hausdorff space.

$$(X,\tau)$$ is Hausdorff if $$\forall x,y\in X\exists U,V\in\tau$$ such that $$x\in U$$ and $$y\in V$$ and $$U\cap V=\emptyset$$.

It easy to note that every Hasudorff space is $$T_1$$-space. However I know that every metrizable space is Hausdorff therefore it is $$T_1$$-space. However I cannot necessarily see or coneceive a proof that a metric space which is both $$T_1$$-space and normal space is Hausdorff space.

Question:

Can anyone help me provide a proof?

Hint: In a $$T_1$$ space points are closed.
If a topological space is normal, then given any two disjoint closed sets $$F_1,F_2$$ there exist disjoint open sets $$U_1,U_2$$ such that $$F_1\subset U_1$$ and $$F_2\subset U_2$$.
If a topological space is $$T_1$$, then any singleton is closed. Therefore, if a topological space is $$T_1$$ and normal, then given any two distinct points $$p_1,p_2$$, the sets $$\{p_1\}$$ and $$\{p_2\}$$ are disjoint closed sets, so by normality, there exist disjoint open sets $$U_1,U_2$$ such that $$p_1\in U_1$$ and $$p_2\in U_2$$, which means that the space is Hausdorff.
There is no need to assume that the space has a metric. In fact, in a metric space, all finite subsets are closed, so in particular, every metric space is $$T_1$$. Also, every metric space is normal, so every metric space is also Hausdorff. Heuristically speaking, if you can measure distance between distinct elements, then you can separate them.