May be, all you need is in the videos on Youtube Amit quoted above, but I would add some concise observations, mainly through examples, that I hope could be useful to you and other contributors.
Let's begin from Point 3 of your question.
Point 3.
To have a simple and intuitive grasp of the concept of quasi-concavity, it could be convenient to see the symmetric concept of quasi-convexity for functions from $\mathbb{R}$ to $\mathbb{R}$.
(Using quasi concavity would be the same, as quasi-concavity is quasi-convexity with the minus sign before. I use quasi-convexity because I think that pictures and formulas are clearer).
The formal definition is: a function $f$ from $\mathbb{R}$ to $\mathbb{R}$ is quasi-convex if, $\forall \alpha \in \mathbb{R}$, the set
$$\{x\in D: f(x)<\alpha\},$$
where $D$ is the domani of $f$, is convex.
In the picture below there is an example of a non quasi-convex function and of a quasi-convex function which is not convex.
In the first picture we have a non quasi-convex function, $f(x)= x^2(x^2-2)$: the set of points for which $f(x)$ is under the green line is the union of the two red intervals, and is not convex.
The second picture shows the function $f(x)=\sqrt|x|$, which is quasi-convex, but not convex.
Point 1.
how can the function be quasi-concave and the plotted curve be convex?
Remember that indifference curves are level curves of a utility function.
A quasi-concave function can ,of course, have convex level curves.
Let's see an example, of a function from $\mathbb{R^2}$ to $\mathbb{R}$.
Consider a Cobb-Douglas utility function, $U(x,y)= kx^\alpha y ^{1-\alpha}$, $x,y\geq0$, $k \in R_+$ and $\alpha \in (0,1)$.
The Cobb Douglas is strictly concave, so it is concave, so it is quasi-concave. The indifference curves are convex.
You can have an intuitive idea from the following picture, where the graphs of the Cobb Douglas in three dimensions, and of its level curves (indifference curves) are represented.
If you calculate analytically the level curves of the Cobb-Douglas, you can see that they are convex.
For example, for $k=1$ and $\alpha=1/2$, we have the hyperbolas
$$ y=c^2/x,$$ $c\in R$, for $x>0$.
Point 2.
Quasi-concavity implies concavity so I can just prove utility is
concave to prove ID curves are convex? Concavity can be proven through
a Hessian matrix of second derivatives right?
I think you made a typo, the reverse of what you wrote is true: concavity implies quasi-concavity (quasi-concavity is a weaker property), so it is sufficient to prove concavity.
Concavity can be proved through second derivatives for functions from $\mathbb{R}$ to $\mathbb{R}$, or through the Hessian matrix, for functions from $\mathbb{R^n}$ to $\mathbb{R}$, but provided that the utility function is differentiable as needed.
For example, one can’t prove the convexity of $f(x))= |x|$ using the derivatives, because this function is not differentiable in all $\mathbb{R}$ ( it is not differentiable in $0$).
