Hoare Calculus Incorrect Assignment Axiom
Your solution is incorrect because you did not reach a contradiction. You claim that you reached a contradiction with the Assignment axiom. For this to be true, you must produce the negation of an instance of the axiom.

The axiom of assignment states that
$$\{P[e/x]\} \; x \mathbin{:=} e \; \{ P \}$$
In your example we would have $u$ instead of $x$, $t$ instead of $e$, and $P$ would be the statement $u = t$, which gives us the following instance of the axiom
$$\{ t = t \} \; u \mathbin{:=} t \; \{ u = t \}$$
You seem to think this is in contradiction with
$$\{ t = u + 1\} \; u \mathbin{:=} t \; \{ u = t \}$$
but that is a logical error. In fact, the axiom of assignment implies your statement:


$\{ t = t \}\; u \mathbin{:=} t \; \{ u = t \}$ (instance of the axiom of assignment, as above)
$\{ t = u + 1 \}\; u \mathbin{:=} t \; \{ u = t \}$ (weakening of the precondition because $t = u + 1 \Rightarrow t = t$.)


Hint: to solve the exercise, consider the possibility that the expression $t$ contains $u$.