Let $S=\{r:[-1,1]\times[-1,1]\subseteq B((0,0);r)\}$. Find $\sup(S)$ and $\inf(S)$.
For $r$ arbitrarily large, the square $[-1,1]\times[-1,1]$ will always be contained in the ball $B((0,0);r)$ so there is no upper bound for $r$, and $\sup (S)=\infty$.
Now for the infimum. In order for the square $[-1,1]\times[-1,1]$ to be a subset of $B((0,0);r)$, the radius of the ball must be at least the norm of the vector $(1,1)$ as that is the furthest point on the square away from the origin. Then $\inf(S) = \sqrt2$.
Is this correct, and have I given sufficient justification?
Sufficient justification usually depends on the rigor of the course. It typically increases as you go through higher levels of mathematics.
Your reasoning regarding the supremum seems fine to me. As for the infimum, the argument needs a little more rigor and your idea is on the right track.
The set $S$ of radii is clearly bounded below by $\sqrt2$ as you argue, thus $inf(S)$ exists (implied by the least upper bound property of real numbers). It now remains to show that $inf(S)= \sqrt2$. Consider the definition of the infimum: $\forall r \in S:$ $\sqrt2 \leq inf(S) \leq r$.
To show $inf(S) = \sqrt2$, why not show $\sqrt2 \geq inf(S)$. One way to do this would be to suppose by way of contradiction that $\sqrt2 <inf(S)$.