A finite graph on at least $k+1$ vertices is $k$-connected if and only if every two vertices have $k$ internally disjoint paths

Suppose $G$ is $k$-connected and $x,y$ are nonadjacent. Every $x$-$y$ separator has at least $k$ vertices by [F1], so [L1] gives $k$ internally disjoint $x$-$y$ paths.

Suppose instead that $xy$ is an edge and that $S$ separates $x$ from $y$ in $G-xy$ with $|S|\le k-2$. If the $x$-component of $(G-xy)-S$ contains a vertex other than $x$, deleting $S\cup\{x\}$ disconnects that vertex from $y$, contrary to [F1].

If that $x$-component is the singleton $\{x\}$, then $N_G(x)\subseteq S\cup\{y\}$, so $|N_G(x)|\le k-1$; deleting $N_G(x)$ leaves the isolated $x$ and another vertex because $|V(G)|\ge k+1$, again contradicting [F1].

Thus every $x$-$y$ separator in $G-xy$ has size at least $k-1$. By [L1], $G-xy$ contains $k-1$ internally disjoint $x$-$y$ paths, and adjoining $xy$ gives $k$ such paths in $G$.

Conversely, suppose every pair has $k$ internally disjoint paths. If $|S|<k$ and $x,y\notin S$, at most $|S|$ of those paths have an internal vertex in $S$, so one survives in $G-S$; as at least two vertices survive, no such $S$ is a vertex cut.

Taking $S=\varnothing$ in step 1.5 gives connectivity, and then [F1] makes the absence of cuts of size below $k$ exactly $\kappa(G)\ge k$; steps 1.1--1.4 prove the converse direction.