Dimension is preserved by normalization How do I show that the dimension of a scheme is the same as its normalization's?

Let $v: \tilde X\rightarrow X$ be the normalization of $X$. I want to show that dim $\tilde X$ = dim $X$.

At first, I wanted to say that the dimensions should be equal because the normalization is a birational map that induces a bijection on the irreducible components. But this doesn't seem to be sufficient because of the following example I thought of: let $\pi: \Bbb A^1 \rightarrow Y$, where $Y$ is a line with a second line glued to its origin, so that the origin is the generic point of the second line. Then we have a birational map sending $\Bbb A^1 \setminus \{0\}$ to the equivalent subset of $Y$, which is dense there as well. However, dim $Y = 3$, so birational maps don't preserve dimension.

For the normalization, we have in addition that $\forall p\in \tilde X$, dim $\mathcal O_{\tilde X, p} =$ dim $\mathcal O_{X, v(p)}$, so dim $\tilde X \le$ dim $X$ But how do I get the other inequality?
I guess a rigorous solution is this:

a normalization gives a bijection between the irreducible closed subsets. Since dimension is determined locally on any irreducible component that maximizes 'the dimension' (more precisely, that maximizes the maximal length of a chain of irreducible closed subsets), we can pick an affine open Spec $A \subset X$ that has that maximal dimension and look at its preimage Spec $B \subset \tilde X$. Then $A$ and $B$ have the same dimension, which gives us the other inequality.
I guess a rigorous solution is this:

a normalization gives a bijection between the irreducible closed subsets. Since dimension is determined locally on any irreducible component that maximizes 'the dimension' (more precisely, that maximizes the maximal length of a chain of irreducible closed subsets), we can pick an affine open Spec $A \subset X$ that has that maximal dimension and look at its preimage Spec $B \subset \tilde X$. Then $A$ and $B$ have the same dimension, which gives us the other inequality.