Show that any metrizable space $X$ is Hausdorff
Solution 1:
Hint: It essentialy boils down to showing that metric spaces are Hausdorff. Take $x,y \in X$ with $x \neq y$. Then $d(x,y) > 0$. Can you check that $$B\left(x,\frac{d(x,y)}{2} \right) \cap B\left(y,\frac{d(x,y)}{2}\right) = \varnothing?$$You should convince yourself with a drawing. To prove it formally, take $z$ in that intersection and get a contradiction using the triangle inequality.