Maschke's theorem states that for a finite group $G$ and a $K$-vector space $V$ under the assumption of $\text{char} K \nmid |G|$, given a representation $\rho: G\to \text{GL}(V)$, every $G$-invariant $W\subset V$ admits a $G$-invariant complement $U$, i.e. $V=U\oplus W$.
I am given the following counterexample for $\text{char} K \mid |G|$: For $G=C_p = \{g^0,g^1,\ldots,g^{p-1}\}$, $K=\mathbb{F}_p$ and $V=K^2$, consider $$\begin{align} \rho: G &\to \text{GL}_2(K)\\\\ g^b &\mapsto \begin{pmatrix} 1 & b\\ 0 & 1 \end{pmatrix}. \end{align}$$ I want to show that no such decomposition is possible in that case. My idea goes as follows:
Assume $V = U\oplus W = \langle (x_1,x_2)^t\rangle \oplus \langle (y_1,y_2)^t\rangle$ and WLOG $x_2\neq 0$. We have for all $g^b\in C_p$ that $$ \rho(g^b)(x_1,x_2)^t = (x_1 + bx_2, x_2).$$ Since $U$ is $G$-invariant, $(x_1 + bx_2, x_2)\in U$, i.e. there is some $\lambda\in \mathbb{F}_p$ such that $$ (x_1 + bx_2, x_2) = (\lambda x_1, \lambda x_2). $$ Since $x_2=0$, the equality in the second component yields $\lambda=1$ and hence $$ x_1 + bx_2 = \lambda x_1 \implies bx_2 = 0. $$ Since this holds for all $b$, we have $x_2=0$, a contradiction.
Now, I do not see where I used the special property of the characteristic $p$ of $\mathbb{F}_p$ in my proof. Any details that I overlooked?