What is the pH of a solution made by mixing 10.00 mL of 0.10 M acetic acid with 10.00 mL of 0.10 M KOH?
Part A:  The conceptually part: Scientific concepts and conception

1-"I thought because 1 mmol of KOH reacts with 1mmol of CH3COOH this will form 1 mmol of CH3COOK"

yes ,the reaction reached at the equivalent point, the weak acetic acid in mixture has been fully neutralized by the stong($\ce{KOH}$) base (no unreacted acid or base present anymore), the solution will only contain $\ce{H2O}$ and the basic salt ($\ce{ CH3COOK}$),so the concentration of the salt :
 $$[\ce{CH3COOK}] =\frac{\text{moles of salt}}{\text{Total volume}} =\frac{1\text{m}mol}{(10+10)\text{ml}}=0.05\text{M}$$ 
 2-"which leads to no formation of OH−ions. Because there is no excess."

No, the result, at the equivalence point, will be the same as dissolving $(\ce{CH3COOK})$ in water at the same concentration(0.05${M}$) as you have at the equivalence point.
the salt ($\ce{ CH3COOK}$) is  electrically neutral substance formed by cation $\ce{K+}$ (an acid)and  anion$\ce{ CH3COO−}$ (a base), completely dissociated or ionized in an aqueous solution as :
  $$\ce{CH3COOK   -&gt; K^+_{(aq)}   + CH3COO^−{(aq)}}$$ 
a)  Since $\ce{K+}$  is the conjugate acid of a strong base, it won't be strong enough to react with water; $\ce{K+}$  actually spectator ion.

b)  Meanwhile, since $\ce{ CH3COO−}$ is the conjugate base of a weak acid, and  therefore strong enough to be able to hydrolyze and accept ions$\ce{ H+}$ from water, so water act as an acid leaving a hydroxide ion $\ce{OH−}$ as :

$$\ce{CH3COO^− +H2O &lt;=&gt;CH3COOH + OH^−}$$

Part B:: Calculating the  approximate pH of $\pu{0.05 mol L-1} \ce{CH3COOK}$ with neglecting water autoionization:

The equilibrium equation of the hydrolysis of the conjugate base $\ce{CH3COO-}$:

$$ K_\mathrm{b}=\frac{K_\mathrm{w}}{K_\mathrm{a}} =\frac{[\ce{OH-}][\ce{CH3COOH}]}{[\ce{CH3COO-}]}$$

Assuming :${[\ce{OH-}]=[\ce{CH3COOH}]}$, ${[\ce{CH3COO-}]=[\ce{CH3COOK}]_0}$

Substiute $K_\mathrm{a}\ and\ [\ce{CH3COO^-}]$ in the equilibrium equation and solve for $[\ce{OH-}]$.

$$[\ce{OH-}] = 5.27\times{10^{-6}} , \mathrm{pOH}=5.278 , \mathrm{pH}=8.72$$