DFS miniumum spanning tree
No. Consider the counterexample $C_4$

$V = \{1,2,3, 4\}, E = \{ \{1,2\}, \{2,3\}, \{3, 4\}, \{1,4\} \}$

with $c(1,2) = 1, c(2,3) = 100, c(3,4) = 1, c(1,4) = 2$.

Obviously the MST is the one ST not containing $\{2,3\}$, but if your algorithm starts at 1, it goes to 2, then to 3 and then to 4 and finds a suboptimal ST containing $\{2,3\}$. 

The main problem of your algorithm is that it always computes a path (i.e if there is no path of length $|V| - 1$ your algorithm can terminate without finding a ST), which the MST must not always be.