Openness of holomorphic maps Chapter II, Theorem 1.18 of Several complex variables VII: Sheaf theoretical methods by Grauert, Peternell and Remmert states:

Let $f: X \to Y$ be holomorphic, and assume that $Y$ is locally irreducible. Then the equation \begin{equation}\dim_x X = \dim_{f(x)}Y + \dim_x X_{f(x)}\end{equation} holds for all points $x \in X$ if and only if the map is open.

Now consider for example a blow-up $f: X \to \mathbb{A}^2$ of $\mathbb{A}^2$ in $0$, i.e. the special fiber over $0$ is a $\mathbb{P}^1$, and $X \setminus \mathbb{P}^1 \to \mathbb{A}^2 \setminus 0$ is an isomorphism. Then for any point $x \in \mathbb{P}^1$, the criterion of the theorem is not satisfied, as $\dim_x X = 2$, $\dim_{f(x)} \mathbb{A}^2 = 2$ and $\dim_x X_{f(x)} = \dim_x \mathbb{P}^1 = 1$.
But I don't believe that $f$ is not open: If $U \subset X$ is open and does not meet $\mathbb{P}^1$, then clearly $f(U)$ is open. So assume $U \cap \mathbb{P}^1 \not = \emptyset $. Then $f(U \setminus \mathbb{P}^1) \subset \mathbb{A}^2 \setminus 0$ is open, and $f(U \cap \mathbb{P}^1) = \{0\}$. So if $x \in U \cap \mathbb{P}^1$, one can choose a small ball $V \subset U$ around $x$, which maps to a small ball in $\mathbb{A}^2$ around $0$. Hence $0$ is contained in the interior of $f(U)$, and so $f$ is open.
What did I not understand here?
 A: The map $f$ is not open.
We can describe $X$ as
$$ X = \left\{ ((x,y),[u,v]) \mid xu=yv \right\} \subset \mathbf A^2 \times \mathbf P^1. $$
The map $f$ is simply the projection $((x,y),[u,v]) \mapsto (x,y)$.
Now consider the open set $U = \{ v \neq 0 \} \subset X$. I claim that $f(U)$ is not an open set of $\mathbf A^2$.
If $(x,y)$ is a point of $\mathbf A^2$ with $x \neq 0$, then setting $u=\frac{y}{x}$ we get that the point $((x,y),[u,1])$ is in $U$, and $f((x,y),[u,1])=(x,y)$.
This shows that the image of the subset $\{x \neq 0\} \subset U$ is exactly the subset $\{x \neq 0\} \subset \mathbf A^2$.
On the other hand, if $((x,y),[u,v]) \in U$ and $x=0$ then we must have $y=0$. So the image of the subset $\{x=0\} \subset U$ is just the point $(0,0)$.
So $f(U) = \{x \neq 0\} \cup \{ (0,0) \}$, which is not open.

To address your actual question, the assertion
"So if $x \in U \cap \mathbf P^1$, one can choose a small ball $V \subset U$ around x, which maps to a small ball in $\mathbf A^2$ around 0"
is incorrect. The image of such a small ball is not a ball in $\mathbf A^2$, as can be seen for example by my description above of the image of $f$.
