Is this theorem of Dold easy to prove with the modern language at hand?

There is a theorem of Dold (in Partitions of unity in the theory of fibrations) saying that if $X$ is a CW-complex and $Y\to X$, $Y'\to X$ are two fibrations connected by a map $f:Y\to Y'$ over $X$ then $f$ is a homotopy equivalence iff $f$ is a fibre homotopy equivalence (meaning that the homotopy inverse is also a map over $X$).

I wonder if this, for a connected $X$ and CW-complexes $Y$ and $Y'$, is an easy consequence of the modern language (of course, the difficulties are just hidden then) as follows:

The model structure on topological spaces $TOP$ induces a model category on the slice category $TOP/X$. Then $Y$ and $Y'$ are cofibrant (as they are CW-complexes) and fibrant(!) objects and $f$ is a morphism in $TOP/X$ which is a weak equivalence. Hence, the Whitehead theorem implies that $f$ is a homotopy equivalence in $TOP/X$ and therefore the inverse is also over $X$.

I think your argument is correct. I assume you are working with Quillen model structure, since you mention $$Y, Y \prime$$ being cofibrant since they are CW-complexes. However, due to this fact the result is not really as strong as we'd want it to be, since very important classes of fibrations do not have domain a CW-complex (say, the path fibration).