I'm not sure what definition of supersingular you're taking; I'll assume you mean that the endomorphism ring is an order in a quaternion algebra.

Now, suppose $E$ is a supersingular elliptic curve over $\mathbb{F}_{q}$ of characteristic $p$, and $\varphi$ is the Frobenius endomorphism. You can deduce from noncommutativity of the endomorphism ring (over the algebraic closure) that $[p^n] = \varphi^m$ for some $m,n \in \mathbb{Z}$. The argument goes along the lines of: if $[p^n]$ is never equal to $\varphi^m$, this forces the endomorphism ring to commute as every endomorphism commutes with some power of the Frobenius. You can find more details in *Elliptic Curves* by Husemoller, including a proof of the converse.

From this, we deduce that multiplication by $p$ is purely inseparable on a supersingular elliptic curve, and hence there is no $p$-torsion.

We can also deduce the following, by degrees of endomorphisms. Let $\alpha,\beta$ be two endomorphisms, then define $\langle \alpha, \beta \rangle = \frac{1}{2} \left( \deg(\alpha + \beta ) - \deg(\alpha) - \deg(\beta) \right)$, this defines an inner product on the endomorphism ring. Using the above property that $[p^n] = \varphi^m$, some basic algebra shows that $p \mid t$, where $t$ is the trace of the Frobenius, which is equal to $\langle \varphi, 1 \rangle = \#E(\mathbb{F}_q) - (q+1)$. It is however not necessarily the case that $t=0$; this happens for $q=p$, $p \geq 5$ by Hasse's inequality, but there are examples in which $t \neq 0$. Going the other way, we see that $\#E(\mathbb{F}_q) \equiv 1 \pmod{p}$, so by elementary group theory $E(\mathbb{F}_q)$ has no $p$-torsion, which is equivalent to supersingularity as shown above.

For the question about reduction modulo $p$, the full criterion of Deuring is as follows:

**Theorem**: Let $L$ be a number field, and $E$ an elliptic curve with complex multiplication by an order in the imaginary quadratic field $K$.

Let $p$ be a (rational) prime, and $P$ a prime above $p$ at which $E$ has good reduction.

Then $E$ has supersingular reduction at $P$ iff there is a unique prime of $K$ above $p$. Otherwise, write $c$ for the conductor of the endomorphism ring of $E$ in $K$, and let $c = c_0 p^k$ with $p \nmid c_0$. Then the ring of endomorphisms of the reduction mod $p$ is $\mathbb{Z} + c_0 \mathcal{O}_K$, the order of $K$ with conductor $c_0$.

I have taken this from Lang's book *Elliptic Functions*, which contains a proof (page 182 in my edition).