$\mathbf{NC}$ is closed under logspace reductions
Given any language $L$ in NL decided by some machine $T$, you can decide whether $x \in L$ by considering the configuration graph of $T$ when run on $x$. This is why PATH is NL-complete. Given any length $n$ of the input, the configuration graph is a fixed polynomial-size graph which doesn't depend on $x$. Only the starting vertex depends on $x$. We can arrange that the starting configuration corresponding to an input $x$ is configuration (vertex) number $x$. This implies that $L$ is in NC.

If we are careful we can do better and show that $L$ is in fact in uniform NC (under various definitions of uniform). The idea now is that given two configurations $\sigma_1,\sigma_2$, it is easy to decide whether $\sigma_1$ can reach $\sigma_2$ in one step. According to the complexity zoo, this way we can show that PATH is NL-complete under first order reductions (uniform AC0 reductions). Since uniform NC is closed under first order reductions, this implies that NL is in uniform NC.