Find an error in computation of $H^n(RP^\infty;\mathbb{Z}_2)$ using cellular structure of the real projective space

Question

I'm following Milnor and Stasheff, Characteristic classes, now it leaves without proof the computation of $H^*(RP^\infty;Z_2)$. I know about the computation in Hatcher's book, nevertheless I thought I could use the CW complex structure that is computed in the book and a couple of theorems to avoid doing such long proof, I tried but I found a contradiction and I cannot locate the error. So if anyone would be so kind to point out were I'm mistaken or how to fix the proof I would appreciate it.
Here are the facts I have already seen:

1. I know that $RP^\infty$ has exactly one cell for each dimension as a CW complex, check corollary 6.5.
2. Since we are working in a field $\mathbb{Z}_2$ then $H^i(X)=Hom(X,\mathbb{Z}_2)$. Check theorem A.1 and apply the fact that the ring is a field.
3. If $K$ is a CW complex and $R$ a commutative ring with unity then the relative homology group $H_i(K^n,K^{n-1})$ with coefficients in $R$ is 0 if $i\neq n$ and is a free $R$-module if $i=n$ with a generator for each $n$-cell of $K^n$, check lemma A.2, in particular if it consists of a single cell one gets it is isomorphic to $R$.
4. If $K$ is a CW complex then $H^i(K^n)$ is zero for $i>n$ and isomorphic to $H_i(K)$ for i<n. The same statement follows for cohomology. Check corollary A.3

Since there is only one generator for each cell in $K^n$ then $H_n(K^n,K^{n-1})$, by applying 3 we get that, it is a free module over $\mathbb{Z}_2$ with rank one which means it is isomorphic to $\mathbb{Z}_2$. Applying 2. we have that $H^n(K^n,K^{n-1})=Hom(\mathbb{Z}_2,\mathbb{Z}_2)=\mathbb{Z}_2$.

Let $K=RP^\infty$ now we compute H^n(K^n) by induction, for $n=0$ by applying 2. $H^0(K^0)=End(\mathbb{Z}_2)=\mathbb{Z}_2=\mathbb{Z}_2[a]/[a]$. Now with the short exact sequence of cochains given by the quocient we derive the long exact sequence: $$ ...\longleftarrow H^n(K^{n-1})\longleftarrow H^{n}(K^n)\longleftarrow H^{n}(K^n,K^{n-1})\longleftarrow H^{n-1}(K^{n-1})\longleftarrow H^{n-1}(K^n)\longleftarrow H^{n-1}(K^n,K^{n-1})\longleftarrow... $$ now we already know that $H^n(K^n,K^{n-1})=\mathbb{Z}_2$, by 2 we also get $H^{n-1}(K^n,K^{n-1})=0$ and by 4 we have that $H^{n}(K^{n-1})=0$ and that $H^{n-1}(K^n)=H^{n-1}(K^{n-1})=\mathbb{Z}_2[a]/[a^n]$ by induction. Thus we are left with the sequence: $$ ...\longleftarrow 0\longleftarrow H^{n}(K^n)\longleftarrow \mathbb{Z}_2\longleftarrow \mathbb{Z}_2[a]/[a^n]\longleftarrow \mathbb{Z}_2[a]/[a^n]\longleftarrow 0\longleftarrow... $$ Now by exactness, the map between $\mathbb{Z}_2[a]/[a^n]$ and the $\mathbb{Z}_2[a]/[a^n]$ is an isomorphism since it is injective because of the 0 and therefore bijective since they are finite modules of the same dimension and it is a homomorphism. Thus the map from $\mathbb{Z}_2[a]/[a^n]$ to $\mathbb{Z}_2$ must be 0, which concludes that $\mathbb{Z}_2\cong H^{n}(K^n)$ but we know this is false, so where is the error?