Surjective morphism of affine varieties and dimension

Let $f:X\to Y$ be your morphism of algebraic varieties over the field $k$.
a) By considering the restricted mapping $f^{-1}(Y_i) \to Y_i$ where $Y_i\subset Y$ is an irreducible component of maximal dimension , you reduce to the case where $Y$ is irreducible .

b) If $X_j$ are the finitely many irreducible components of $X$, you have $Y=\bigcup f(X_j)$, hence $Y=\bigcup \overline {f(X_j)}$, so that one of the $f(X_j)$ is dense in $Y$.
In other words by considering $f_{\big|X_j}: X_j\to Y $ you reduce to the case that $X$ is irreducible but you must weaken yout hypothesis to $f$ is dominant (that is with dense image) instead of $f$ is surjective. Don't worry : we'll still manage!

c) Since the morphism $f:X\to Y$ of irreducible varieties is dominant, it induces a morphism of the corresponding function fields $f^* :\operatorname{Rat}(Y) \to \operatorname{Rat}(X) $ .
This implies by pure algebra (field theory!) the inequality on transcendence degrees over $k$ : $$\operatorname{trdeg}_k (\operatorname{Rat}(Y))\leq \operatorname{trdeg}_k (\operatorname{Rat}(Y)) \quad (*)$$
d) Finally, for an irreducible variety $Z$, we know that $\dim (Z)=\operatorname{trdeg}_k (\operatorname{Rat} (Z)$ (this is essentially a corollary of Noether's normalization theorem). If you take this into account, $(*)$ yields the required inequality $$ \dim(Y)\leq \dim(X) \quad (**) $$

Edit The above works for all algebraic varieties, affine or not. Actually I hadn't even noticed that the OP mentioned affine varieties when I answered the question!