Similar versions of this question has been posted before, but a complete solution was never posted from what I can see. I have an attempt at a solution which I'd like to have checked for errors.
I have a graph $G$ with $n$ vertices, and the probability of any two vertices having an edge is $p \geq \frac{\ln n}{n-1}$.
Now, take some subset $G_i$ of the graph $G$ and let $g_i$ be the number of vertices in $G_i$. I want to prove that the likelihood of there being no edge connecting a vertex in $G_i$ to a vertex outside $G_i$ is very unlikely.
Firstly, consider the case $g_i = 1$.
We have that $$P[\:\nexists \text{ an edge (u,v) for u,v}\in G] = (1-p)$$ Therefore since $g_i = 1$, $$P[\:\nexists \text{ an edge from } G_i \text{ to } G_i^c] = (1-p)^{(n-1)}$$
Now let $p = (1+\epsilon)\frac{\ln n}{n-1}$ where $\epsilon > 0$. Then $n-1 = \frac{1}{p}(1+\epsilon)\ln n$ so
$$(1-p)^{(n-1)} = (1-p)^{\frac{1}{p}(1+\epsilon)\ln n}$$
And now using the fact that in general for positive $p$, $(1-p)^{1/p} \leq 1/e$ I have that
$$(1-p)^{\frac{1}{p}(1+\epsilon)\ln n} \leq \frac{1}{e^{(1+\epsilon)\ln n}} = \frac{1}{n^\epsilon}$$
Therefore for $g_i = 1$ (an isolated vertex), the probability vanishes as $p$ increases further past the threshold of $\frac{\ln n}{n-1}$
Further, if $g_i = n$ then of course the graph is connected, so let $2 \leq g_i \leq n-1$. Therefore since we're comparing every vertex in $G_1$ to every vertex outside it,
$$P[\:\nexists \text{ an edge from } G_i \text{ to } G_i^c] = (1-p)^{g_i(n-g_i)}$$
Now I want to find the maximum value of $g_i$ for which the above probability is maximised. That is, where $y = g_i(n-g_i)$ is minimised. The expression $y$ can be thought of as a parabola where the domain is $g_i \in [2,n-1]$. It is quick to check that the minimum point occurs at $g_i = n-1$, therefore I have that
$$P[\:\nexists \text{ an edge from } G_i \text{ to } G_i^c] \leq (1-p)^{(n-1)(n-(n-1))} = (1-p)^{(n-1)}$$
But $(1-p)^{(n-1)}$ is the probability in the case of $g_i = 1$, which has already been shown to vanish as $p$ exceeds $\frac{\ln n}{n-1}$. Therefore the probability of any subset of $G$ being disconnected also vanishes.