The problem with your proof is where you say "until you obtain $w$".  That makes it sound to me like you stop the search as soon as you find a single $x$ such that $f(x)=w$.  If that is what your machine $M'$ does, then your proof is faulty, for [the reasons chi explains](https://cs.stackexchange.com/a/87710/755).

Also, there is an additional step that is worth explaining: it's worth mentioning that any language that's recognizable by a NTM is also recognizable by a standard deterministic TM.  This is a [standard fact](https://en.wikipedia.org/wiki/Non-deterministic_Turing_machine#DTM_simulation_of_NTM) that probably doesn't need to be proven here.

Fortunately, there is a simple fix.  Here is an improved proof:


> Given a language $L$ that is Turing recognizable and a TM $M$ that recognizes it and a homomorphism $f$, we build a NTM $M'$ that recognizes $f(L)$.

> $M'$ looks like this:

> On input $w$ :

> - Non-deterministicly guess a word $x \in \Sigma^*$.

> - If $f(x)=w$ and $M$ accepts on input $x$, accept, otherwise reject.
> 
> This works because $M'$ is a non-deterministic Turing machine.  A NTM accepts $w$ if there is at least one branch accepting $w$, so if there is any word $x$ such that $f(x)=w$ and $M(x)$ accepts, this will find it and accept.
> 
> Moreover, any language that can be recognized by a nondeterministic TM, [can be](https://en.wikipedia.org/wiki/Non-deterministic_Turing_machine#DTM_simulation_of_NTM) recognized by a deterministic TM.  It follows that $f(L)$ is Turing recognizable.

Credit: thanks to [xskxzr](https://cs.stackexchange.com/a/89201/755) for explaining the idea of the proof and for the improved formatting.


  [1]: https://cs.stackexchange.com/q/87709/83244