At the start of your derivation you multiply out the brackets ∑i(xi−x¯)(yi−y¯), in the process expanding both yi and y¯. The former depends on the sum variable i, whereas the latter doesn't. If you leave y¯ as is, the derivation is a lot simpler, because
∑i(xi−x¯)y¯=y¯∑i(xi−x¯)=y¯((∑ixi)−nx¯)=y¯(nx¯−nx¯)=0
Hence
∑i(xi−x¯)(yi−y¯)=∑i(xi−x¯)yi−∑i(xi−x¯)y¯=∑i(xi−x¯)yi=∑i(xi−x¯)(β0+β1xi+ui)
and
Var(β1^)=Var(∑i(xi−x¯)(yi−y¯)∑i(xi−x¯)2)=Var(∑i(xi−x¯)(β0+β1xi+ui)∑i(xi−x¯)2),substituting in the above=Var(∑i(xi−x¯)ui∑i(xi−x¯)2),noting only ui is a random variable=∑i(xi−x¯)2Var(ui)(∑i(xi−x¯)2)2,independence of ui and, Var(kX)=k2Var(X)=σ2∑i(xi−x¯)2
which is the result you want.
As a side note, I spent a long time trying to find an error in your derivation. In the end I decided that discretion was the better part of valour and it was best to try the simpler approach. However for the record I wasn't sure that this step was justified
=.1(∑i(xi−x¯)2)2E⎡⎣(∑i(xi−x¯)(ui−∑jujn))2⎤⎦=1(∑i(xi−x¯)2)2E[∑i(xi−x¯)2(ui−∑jujn)2] , since ui 's are iid
because it misses out the cross terms due to
∑jujn.