Outer Hausdorff measure/Infinite Let $\mathcal{H^s}$ be the s-dimensional outer Hausdorff measure. 

For all $t \in (0,n)$ there is none countable cover $\lbrace F_i \rbrace^\infty_{i=1}$ 

of $\mathbb{R^n}=\bigcup_{i=1}^{\infty} F_{i}$ with $\mathcal{H^s}(F_i)&lt;\infty$

I used the contradiction::

If it would be possible then

$n=$dim$_H(\mathbb{R^n})=$sup dim$_H(F_i)\leq s&lt;n$ 

So there can't be such a cover. 

Can it be proved like this or is there something to improve?
That looks fine, provided that you can prove $$\dim_H \left( \bigcup_i F_i \right) = \sup_i \dim_H(F_i).$$ It shouldn't be a difficult proof.
That looks fine, provided that you can prove $$\dim_H \left( \bigcup_i F_i \right) = \sup_i \dim_H(F_i).$$ It shouldn't be a difficult proof.