Center of the Quaternions: Proof and Method

I am not sure where the contradiction lies exactly in your proof by contradiction. But here is another method.

An element $x\in \mathbb H$ belongs to the center if and only if $[x,y]=0$ for all $y\in \mathbb H$, where $[x,y]=xy-yx$ denotes the commutator of two elements.

We see immediately that $[x,1]=0$, whereas if $x=a+bi+cj+dk$ we have $$ [x,i]=-2ck+2dj. $$ Thus $[x,i]=0$ if and only if $c=d=0$. Similarly $[x,j]=0$ if and only if $b=d=0$. Thus the only elements $x$ which commute with both $i$ and $j$ are $x\in \mathbb R$; in particular, it follows that $Z(\mathbb H)\subset \mathbb R$. Since it is clear that $\mathbb R\subset Z(\mathbb H)$, the result follows.

Idea behind the proof: There are three special copies of the complex numbers sitting inside $\mathbb H$: the subspaces $$ \mathbb C_i=\mathbb R[i],\qquad \mathbb C_j=\mathbb R[j],\qquad \mathbb C_k=\mathbb R[k]. $$ Over $\mathbb H$, all of these subspaces are their own centers: $Z_{\mathbb H}(\mathbb C_i)=\mathbb C_i$ and so forth. Since $$\mathbb H=\mathbb C_i+ \mathbb C_j+ \mathbb C_k,$$ it follows that $Z(\mathbb H)=Z(\mathbb C_i)\cap Z(\mathbb C_j)\cap Z(\mathbb C_k)=\mathbb R$.

If $a+bi+cj+dk$ is in center, then it should commute with generators $$i,j,k,\mbox{ and reals}.$$ For example, see what do we get for $(a+bi+cj+dk).i=i.(a+bi+cj+dk)$?