Prove there's a simple path of length $k$ in a simple graph $G$ where all the vertices have degree of at least $k$

The problem with that approach is that the path of length $k-1$ in $G-v_0$ may not have a vertex adjacent to $v_0$ at either end, so you may not be able to extend it to a path of length $k$ in $G$.

You don’t actually need induction on $k$. Start at any vertex $v_0$. It has degree $k$, so it’s adjacent to some vertex $v_1$. Assuming that $k>1$, $v_1$ is adjacent to at least one vertex $v_2$ other than $v_0$. If $k>2$, then $v_2$ is adjacent to at least one vertex $v_3$ different from $v_0$ and $v_1$. And so on. In general you can continue this argument until you’ve chosen $v_k$ adjacent to $v_{k-1}$ and different from $v_0,\ldots,v_{k-2}$, at which point you have a path of length $k$.

Suppose the length of the longest path is $m<k$ and take a look at one of the paths (say it's made up of vertices $v_1, v_2,\dots,v_{m+1}$).

Consider $v_{m+1}$. It cannot be connected to vertices outside of the path, because then we could form a longer path and contradict maximality. Hence it can only be connected to some of the other $v_i$ (given it is a simple graph). But there are only $m<k$ of those, so this contradicts the minimum degree condition of the graph.