ok so i think i got it but not sure.. sorry about the format also, wasn't sure how else to insert it:

Let ∑_(i=1)^n▒〖(x_i-x ̅ )^2=SST_x 〗

Prove β ̂_1=β_1+∑_(i=1)^n▒〖w_i u_i 〗 where; w_i=d_i/(SST_x ) , d_i=x_i-x ̅

β ̂_1=β_1+(∑_(i=1)^n▒〖(x_i-x ̅)u_i 〗)/(SST_x )

=β_1+(∑_(i=1)^n▒〖d_i u_i 〗)/(SST_x )

=β_1 ∑_(i=1)^n▒〖w_i u_i 〗

Prove E[(β ̂_1-β_1 ) u ̅ ]=0 to show β ̂_1 and u ̅ are uncorrelated:

E[(β ̂_1-β_1 ) u ̅ ]=E[u ̅ ∑_(i=1)^n▒〖w_i u_i 〗]

=∑_(i=1)^n▒〖〖E(w〗_i 〖u ̅ u〗_i 〗)

=∑_(i=1)^n▒〖w_i E(〖u ̅ u〗_i 〗)

=1/n ∑_(i=1)^n▒〖w_i E(〖 u〗_i ∑▒u_j 〗)

=1/n ∑_(i=1)^n▒〖w_i [E(〖 u〗_i 〖 u〗_1 )+⋯E(〖 u〗_i 〖 u〗_j )+⋯+E(〖 u〗_i 〖 u〗_n )]〗

=1/n ∑_(i=1)^n▒〖w_i E(u^2 〗)

=1/n ∑_(i=1)^n▒〖w_i [var(〖 u〗_i )+E(〖 u〗_i )E(〖 u〗_i )]〗

=1/n ∑_(i=1)^n▒〖w_i σ^2 〗

=σ^2/n ∑_(i=1)^n▒w_i

=σ^2/(n . SST_x ) ∑_(i=1)^n▒〖(x_i-x ̅)〗

=σ^2/(n . SST_x )

Prove β ̂_0=β_0+u ̅-x ̅(β ̂_1-β_1)

β ̂_0=y ̅-β ̂_1 x ̅

=(β_0+β_1 x ̅+u ̅ )-β ̂_1 x ̅

=β_0+u ̅-x ̅(β ̂_1-β_1)

Prove 〖var(β ̂_0 )=σ^2/n+(σ^2 (〖x ̅)〗^2)/〖SST〗_x 〗_

var(β ̂_0 )=var[β_0+u ̅-x ̅(β ̂_1-β_1 )]

=var(u ̅ )+(-u ̅ )^2 var(β ̂_1-β_1)

=(σ^2/n)+(x ̅ )^2 var(β ̂_1)

=(σ^2/n)+(σ^2 (x ̅ )^2)/〖SST〗_x

Therefore;

var(β ̂_0 )=(σ^2/n)+(σ^2 (x ̅ )^2)/〖SST〗_x

=(σ^2.〖SST〗_x)/(〖SST〗_x.n)+(σ^2 (x ̅ )^2)/SST

=σ^2/〖SST〗_x (1/n ∑_(i=1)^n▒〖x_i^2-(x ̅ )^2)+(σ^2 (x ̅ )^2)/〖SST〗_x 〗

=(σ^2 n^(-1) ∑_(i=1)^n▒x_i^2 )/(∑_(i=1)^n▒(x_i-x ̅ )^2 )