Consider the $\mathbb{Z}/2$-cohomological Serre spectral sequence associated to the path fibration $$ K(\mathbb{Z}/2, 1) \simeq \Omega K(\mathbb{Z}/2, 2) \longrightarrow PK(\mathbb{Z}/2, 2) \longrightarrow K(\mathbb{Z}/2, 2). $$
By universal coefficients, we are allowed to write $E^{p, q}_2 \cong H^p(K(\mathbb{Z}/2, 2); \mathbb{Z}/2)) \otimes H^q(\mathbb{RP}^{\infty}; \mathbb{Z}/2)$. Write $x\in H^1(\mathbb{RP}^{\infty}; \mathbb{Z}/2)$ for the generator of the polynomial algebra $H^*(\mathbb{RP}^{\infty}; \mathbb{Z}/2)\cong \mathbb{Z}/2[x]$ and $\iota_2\in H^2(K(\mathbb{Z}/2, 2); \mathbb{Z}/2)$ for the generator.
Since $PK(\mathbb{Z}/2, 2)$ is contractible, no entry on the $E_2$ page except for $E_2^{0, 0}$ can survive to $E_\infty$. Since $H^1(K(\mathbb{Z}/2, 2); \mathbb{Z}/2) = 0$, we have $E_2^{1, 1} = 0$, and hence the only way that $E_2^{0, 3}$, which I know is generated by $\operatorname{Sq}^1(\iota_2)\otimes 1 \neq 0$, can die is that the map $$d_3\colon E_3^{0, 2}\to E_3^{3, 0}$$ is an isomorphism.
Now it is not hard to see that $d_2\colon E_2^{0, 2}\to E_2^{2, 1}$ must be the zero map, so $E_3^{0, 2}$ is still generated by $1\otimes x^2$. But now we can calculate using the Leibniz rule that (recall that we're working over $\mathbb{Z}/2$, hence signs can be ignored) \begin{align*} d_3(1\otimes x^2) &= d_3((1\smile 1)\otimes (x\smile x))\\ &= d_3((1\otimes x)\cdot (1\otimes x))\\ &= d_3(1\otimes x)\cdot (1\otimes x) + (1\otimes x) \cdot d_3(1\otimes x). \end{align*} But $d_3(1\otimes x)\in E_3^{3, -1} = 0$, and so the entire final term is $0$. This is a contradiction because we argued earlier that the map is an isomorphism.
My takeaway is that there must be an error in the explicit calculation of the differential, but I can't find it. Where is it?
You missed $d_2: E^{0,1} \to E^{2,0}$. This must be an isomorphism, as it is the last chance to kill off $E^{0,1}$. One finds that $d_2(1 \otimes x) = \iota_2 \otimes 1$.
The Leibniz rule for $d_3$ states that if $x,y \in E^3$, then $d_3(xy) = d_3(x)y + (-1)^{|x|} x d_3(y).$
Now that $E_3^{0,1} = 0$ the element $(1 \otimes x)$ does not exist in the $E_3$ page. In particular, your expression $(1 \otimes x)^2 = 1 \otimes x^2$ takes place in the $E_2$ page! You have a Leibniz rule for $d_2$ of this element and indeed as you claim $d_2(1 \otimes x^2) = 0$. But you get nothing for $d_3$.
Indeed $d_3(1 \otimes x^2) = \text{Sq}^1 \iota_2$, just like you say.
It might be interesting to ask what the Leibniz rule ought to say in your situation: for elements $x,y$ which were cycles on previous pages but are no longer. The $\text{Sq}^1$ suggests that this is the home of primary cohomology operations on spectral sequences. I do not know anything about this.