It is shown in this thread that $\dfrac{1}{2}\sqrt{\dfrac{n}{k}}$ is an upper bound for the radius of $k$-dimensional balls that can be contained in an $n$ dimensional unit hypercube.
But I have difficulty seeing that this radius can be attained for any $n,k$. I can draw pictures to see it geometrically for some special cases but not for general high dimensional cases. Can someone show me an argument, constructive or not, to see why this maximum can be attained? (please make sure to verify the ball you find is really contained in the cube)
Edit: It seems you are asking why the specific bound $\displaystyle\frac{1}{2}\sqrt{\frac{n}{k}}$ can be attained, which I have no idea about. The argument below shows that the supremum of radius, whatever it is, must be attained.
That the supremum is attained should be an easy compactness argument once you parametrize $k$-dimensional balls cleverly, for example a disk in $\mathbb{R}^3$ is determined by its radius, center and the axis (the perpendicular line through the center), which can be viewed as an element in the projective plane $\mathbb{P}^2$. I'm not clever enough to generalize this to higher dimension, so let me give a very indirect argument using hyperspace, which is a bit tedius but intuitive in my opinion.
Say $(X,d_X)$ is any metric space. Denote by $K(X)$ the collection of compact subsets of $X$. The Hausdorff distance, which measures how close two compact sets are to each other, makes $K(X)$ into a metric space; the topology induced by this metric also coincides with the Vietoris topology (maybe under some mild condition). It can be proven that if $X$ is compact then so is $K(X)$. Moreover, the assignment to each $F\in K(X)$ of its diameter $\mathrm{diam}(F)=\sup\{d_X(x,y)\mid x,y\in F\}$ is continuous w.r.t. the above topology.
Now for fixed $k$ and $R>0$, consider $H=\{F\in K(X)\mid F\text{ is isometric to a $k$-dimensional ball of radius} \leq R\}$, where we agree that a point is a ball of radius zero. We claim that this set is closed in $K(X)$. Suppose $(F_n)_n\subseteq H$ converges to $F\in K(X)$ under Hausdorff distance. Let $B$ be the standard $k$-dimensional unit ball. If for each $n$, $F_n$ is a ball of radius $r_n$, then there is a map $f_n:B\rightarrow F_n$ which is the composition of a scaling of $B$ by $r_n$ and an isometry with $F_n$; in particular $f_n$ is $r_n$-Lipschitz. Our definition of $H$ implies that $r_n\leq R$, so $(f_n)_n$ is uniformly equicontinuous; they are also uniformly bounded since $(F_n)_n$ converges. By Arzela-Ascoli Theorem, there is a subsequence of $(f_n)_n$ that converges uniformly to some $f$. It follows from the definition of uniform convergence that $f_n(B)\rightarrow f(B)$ in Hausdorff distance, so $f(B)=F$.
From the convergence of $(F_n)_n$ we also see that $\mathrm{diam}(F_n)\rightarrow\mathrm{diam}(F)$, but $\mathrm{diam}(F_n)=2r_n$ (for a ball the diameter in metric sense agrees with the diameter in usual sense), so either $F$ is a point and $r_n\rightarrow 0$, or $r_n\rightarrow r$ for some $r>0$. In the latter case $F$ must be a ball of radius $r$: since $d_X(f_n(a),f_n(b))=r_n\cdot d(a,b)$, where $d$ means the Euclidean distance in $\mathbb{R}^k$, taking limit we have $d_X(f(a),f(b))=r\cdot d(a,b)$, so $f$ is scaling by $r$ followed by an isometry.
Finally, we take $X=[0,1]^n$ and notice that an obvious bound for radius of $k$-dimensional balls in $X$ is $\sqrt{n}$, so if we take $R=\sqrt{n}$ then $H$ contains all $k$-dimensional balls in $X$. Now since $K(X)$ is compact and $H$ is closed, $H$ is also compact, and the diameter function attains supremum on $H$ (again the metric diameter equals the usual diameter of a ball).