What happens as $n \rightarrow \infty$? To answer we require further assumptions! Now assumptions 7 and 8…
7. The regressors and errors are uncorrelated in the limit (as we assumed for the finite sample)
$$\plim_{n\rightarrow \infty} \biggl(\dfrac{1}{n} X’ u \biggr) = \plim_{n\rightarrow \infty} \biggl(\dfrac{1}{n} \sum_{t=1}^n x_t u_t \biggr) = 0$$
8. In the limit, there is no multi-collinearity. (As we assumed for the finite sample case)
$$\plim_{n\rightarrow \infty} \biggl(\dfrac{1}{n} X’ X \biggr) = \plim_{n\rightarrow \infty} \biggl(\dfrac{1}{n} \sum_{t=1}^n x_t x_t’ \biggr) = Q; \text{(p.d.)}$$
$\Rightarrow$ as $n \rightarrow \infty$, no infinities and no singularities. Given that
\begin{align*}
\widehat{\beta} &= (X’X)^{-1} X’ y = \beta + (X’X)^{-1}X’u\\
\Rightarrow \quad \plim_{n\rightarrow \infty} \widehat{\beta} &= \plim_{n\rightarrow \infty} (\beta + (X’X)^{-1}X’u)\end{align*}
By the continuous mapping theorem,\begin{align*}
&= \plim_{n\rightarrow \infty} \beta + \plim_{n\rightarrow \infty} \biggl[ \biggl(\dfrac{1}{n} X’X \biggr)^{-1} \biggl(\dfrac{1}{n} X’u \biggr) \biggr]
\\
&= \beta + \underbrace{\plim_{n\rightarrow \infty} \biggl(\dfrac{1}{n} X’X \biggr)^{-1}}_{Q^{-1}}\times \underbrace{\plim_{n\rightarrow \infty} \biggl(\dfrac{1}{n} X’u \biggr)}_{0}\end{align*}
note the $\dfrac{1}{n}$ cancel out.
\begin{align*}\widehat{\beta} =\beta + Q^{-1}\times 0 = \widehat{\beta}\quad \therefore \quad \text{consistent!}
\end{align*}