Sorgenfrey line - what is wrong with this argument

Question

Consider $X$ the reals with Euclidean topology, $Y$ the reals with Sorgenfrey topology, $f: X \rightarrow Y$ identity mapping $f(x) = x$.

I know that this mapping is not continuous, because $U = [0,1)$ is open in $Y$, but $f^{-1}(U) = [0,1)$ is not open in $X$.

But if we look at the definition of continuity at a point $x_0 \in X$ ($f$ is cts in $x_0 \in X$ iff $\forall V$ neighbourhood of $f(x_0)$ there exists $U$ a neighbourhood of $x_0$ such that $f(U) \subset V$).

So choose a neighbourhood of $f(x_0) = x_0$ in $Y$, say $V = [a,b)$. Now we need to find a neighbourhood $U$ in $X$ of the point $x_0$. Denote $U = (a,b)$. Then $U$ is an open neighbourhood of $x_0$ in $X$ and $f(U) = U = (a,b) \subset [a,b) = V$.

What is wrong with this argument?

Answer 1

It does not work for every neighborhood of $x_0$. In particular, it doesn't work if you take $V=[x_0,x_0+1)$.

Answer 2

The problem happens if $a=x_0$. In that case, $[a,b)$ is still a neighborhood of $x_0$, but $(a,b)$ is not a neighborhood of $x_0$.