Inspired by MSE/95760 I'm wondering whether the following is true: Let $\varphi : V \to W$ be an injective morphism between (affine) varieties. Does it follow that $\dim{V} \leq \dim{W}$?
I am not able to construct any counterexamples and I cannot see a way to translate chains of irreducible varieties in $V$ to $W$.
Let $f\colon X\to Y$ be an injective continuous map of sober topological spaces. Then $\dim(X)\leq\dim(Y)$, for it maps a sequence of non-trivial specialisations $x_n\leadsto x_{n-1}\leadsto\dots\leadsto x_0$ in $ X$, i.e., a sequence of distinct points such that $x_{i-1}\in\overline{\{x_i\}}$ for all $i=1,\dots,n$, to a sequence of non-trivial specialisations $f(x_n)\leadsto f(x_{n-1})\leadsto\dots\leadsto f(x_0)$ in $Y$.
In a bit more detail and without that terminology, if $X_0\subsetneq X_1\subsetneq \dots\subsetneq X_n$ is a chain of proper inclusions of irreducible closed sub-sets, then we get a corresponding sequence of generic points $x_i\in X_i$—i.e., $X_i=\overline{\{x_i\}}$ for all $i=0,\dots,n$—and so $x_{i-1}\in X_{i-1}\subsetneq X_i = \overline{\{x_i\}}$ for all $i=1,\dots,n$. Clearly, the $x_i$ are pair-wise distinct; in fact, $x_i\not\in X_{i-1}$. Consider $y_i:=f(x_i)$. Since $f$ is injective, those points are pair-wise distinct as well and since $Y$ is sober, so are their corresponding components $Y_i:=\overline{\{y_i\}}\subset Y$. Therefore, $Y_0\subsetneq Y_1\subsetneq \dots\subsetneq Y_n$. This is what's been missing to complete the proof attempted in the other answer.
Towards the follow up question in the OPs answer, cf. Stacks Project Tag 02JF.
Someone posted an answer here but unfortunately deleted it because it was flawed. That is a pity because I believe that every mistake is educational and there were also several helpful comments. (Should it really be this easy to just delete answers on MSE?) That's why I am writing this answer. I want to retrieve everything that seems of importance for people in the future and let the previous author stay anonymous.
The claim was that the problem can be solved in a purely topological setting, i.e. let $f:X \to Y$ be an injective continuous map between topological spaces, then $\dim{X} \leq \dim{Y}$ for their Krull dimensions. We then try to lift a chain from $X$ to $Y$, that is let $$X_0 \subsetneq X_1 \subsetneq \dots \subsetneq X_n = X$$ be a chain of irreducible closed sets and look at $$\overline{f(X_0)} \subseteq \overline{f(X_1)} \subseteq \dots \subseteq \overline{f(X_n)}.$$ One can check that this is again a chain of irreducible closed sets and that's where the previous answer finished his proof.
There is a problem. It is not entirely clear that the inclusions in this new chain remain proper. In fact, it is wrong.
Let $X$ be any space of positive Krull dimension. Take any set $Y$ such that there exists an injective function $X \to Y$ and give $Y$ the indiscrete topology. So $\dim{Y} = 0$ and the map is continuous. That is a counterexample.
This example shows that the problem cannot be solved in a purely topological setting and requires some algebraic geometry.
The natural follow-up question is whether the dual problem (again, see MSE/95670) can be solved in a topological setting, i.e. given a surjective continuous map $f:X \to Y$ between topological spaces, do we get $\dim{X} \geq \dim{Y}$?
The answer is no, our previous idea can be applied again. Take a surjective map from a discrete topological space.