If Degree of a Vertex Greater Than/Equal to Size of Vertex Set / 2 then Connected

Question

I encountered a question asking to prove that if the degree of every vertex is greater than or equal to the floor of the size of its vertex set divided by 2, then its connected.
I was initially thinking of doing a proof by cases where I first do where the degree is equal to size of vertex set divided by 2 and the second set when its strictly greater but it doesn't seem to be going anywhere.
Could someone provide some guidance on how to approach this question? Thanks!

Answer 1

Every component of the graph has strictly more than $\frac{\nu}{2}$ vertices in it. Consider one component of the graph. There aren't enough "leftover" vertices left to form a second component. Therefore, there is only one component and the graph is connected.

Answer 2

Take any two non-adjacent vertices $x,y$. You know$$\deg(x)+\deg(y)\ge\left\lfloor\frac v2\right\rfloor+\left\lfloor\frac v2\right\rfloor\ge v-1$$There are $v-2$ vertices other than $x$ and $y$. This means $x$ and $y$ have at least one common neighbour vertex, assuming the graph is simple, i.e. any pair of non-adjacent vertices has a common neighbour, making the graph connected. In addition, the eccentricity of each vertex, radius and diameter of the graph are $\le2$.