"Given an algorithm, decide whether it runs in polynomial time" is this problem in NP? This problem is not decidable (reducible to halting problem) but is semi-decidable and therefor verifiable (as those two definitions are equivalent: How to prove semi-decidable = verifiable?).
However, is this problem poly-time verifiable?
A decision problem $P$ is poly time verifiable iff there is an algorithm  called verifier such that if $P(w)=$ then there is a string $c$ s.t. $(w,c)=$, if $P(w)=$ then for all strings $c$, $(w,c)=$ and V runs in $O(w^{k})$ for some constant $k$ for all inputs $w$.
No.  If the problem was polynomial-time verifiable, it would be solvable in exponential time, and thus decidable; but we already know that is not decidable.
Why in exponential time?  Because $V$ runs in time $|w|^k$, it can read at most $|w|^k$ bits of the input.  So, it suffices to enumerate all possible strings $c$ of length at most $|w|^k$, and run $V$ on each of them.  The running time will be about $2^{|w|^k}$, which is finite and thus enough to make the original problem decidable.