This is more of a long comment than an answer. The approach you are suggesting seems unlikely to work to me. First, let's consider whether the various relevant notions are absolute. Recall that a property is upwards absolute if it is preserved under passing to bigger models of $\mathsf{ZFC}$ with the same ordinals (e.g., forcing extensions) and is downwards absolute if it is preserved under passing to smaller models of $\mathsf{ZFC}$ with the same ordinals (e.g., inner models like Gödel's $L$).
Being a free abelian group and being discretely embeddable in a normed space are both upwards absolute: If $A$ is freely generated by $B \subseteq A$, it will still be freely generated by $B$ in any larger model of $\mathsf{ZFC}$. Likewise if $f: A \to X$ is an embedding into a normed space with discrete image, $f$ will still be an embedding into $X$ and its image will still be discrete. (One thing to note is that $X$ may fail to be a complete normed space in the larger universe, but this isn't really an issue.) The formal way set theorists would talk about this is that both of these properties are $\Sigma^1_1$ (i.e., they're of the form 'there exists a set $Y$ such that [something involving only bounded quantifiers happens]'). Any such property is upwards absolute.
It's not clear to me whether either of them are downwards absolute (because the $B$ or $(f,X)$ witnessing the relevant property may fail to exist in the smaller universe), but I suspect that they probably both are not.
Whiteheadness is not clearly either upwards or downwards absolute. $A$ is whitehead if and only if every exact sequence $0 \to \mathbb{Z} \to B \to A \to 0$ of abelian groups splits. This is a $\Pi^1_2$ statement, something of the form 'for every set $W$, there exists a set $U$ such that [something involving only bounded quantifiers happens].' In this case $W$ is a set coding $B$ and the morphisms in the exact sequence, and $U$ is the morphism witnessing that it splits. (Note that the quantifier for $U$ doesn't 'count' as bounded for our purposes because it's quantifying over the power set of $A\times B$ and new subsets of $A \times B$ may show up in a larger universe.)
We can actually show that Whiteheadness is not upwards absolute explicitly. If we look at Eklof's exposition of the independence of the Whitehead problem, we can see that the non-free Whitehead group constructed in the proof of Theorem 7.3 can actually be constructed in $L$. In other words we can run the construction without assuming $\mathsf{MA} + 2^{\aleph_0} > \aleph_1$ and produce some group $A$, but this group won't be Whitehead in $L$. As is mentioned in this question, there are c.c.c. forcings that force $\mathsf{MA} + 2^{\aleph_0} > \aleph_1$. Since these forcings are c.c.c., they don't collapse $\aleph_1$. So if $L[G]$ is some such forcing extension, we'll have that $A$ is Whitehead in $L[G]$.
I suspect Whiteheadness is also not downwards absolute, but I don't have an argument.
Now to actually address your question, if we're in a model $M$ of $\mathsf{ZFC}$ in which all Whitehead groups are free, any Whithead group $A$ in $M$ will still be free in any extension $N \supseteq M$, so we can't use this to build an example of a discrete subgroup of a normed space that is a non-free Whitehead group.
In principle one could build a non-free discrete subgroup $A$ of some normed space $X$ and then find a forcing extension in which $A$ is Whitehead, but I'm not sure that structuring the argument this way would be necessary. It would in some sense be equivalent to starting with the forcing extension that ensures the existence of non-free Whitehead groups and then build $A$ and $X$ there.
If freeness is not downwards absolute, you might be able to do something by carefully passing to an inner model as well, but you'd need to somehow arrange so that in the inner model the group you built is still Whitehead and still discretely embeddable in a normed space, which seems difficult to me.
