Consider an arbitrary but fixed solution. Let p denote the number of nodes that are lit up at time 0. Let T denote the time taken to lit the entire grid. WLOG, assume that for each 1 \leq t \leq T, exactly one new node is lit up at time t.

At any time 0 \leq t \leq T,

For every lit node (i,j), let f(i,j,t) denote the number of edges of the node (i,j) that are not shared with another lit node.

Let g(t) denote the sum of f(i,j,t) over all lit nodes (i,j)

We show that g(t+1) \leq g(t) for all 0 \leq t \leq T-1.

Let 0 \leq t \leq T-1. Let (u,v) denote the new node which is lit up at time t+1. Let X denote the set of all lit neighbors of (u,v) at time t. For every node (i,j) in X, we have f(i,j,t+1) = f(i,j,t) - 1. Also, note that f(u,v,t+1) = 4-|X|. Therefore, g(t+1) = g(t) - |X| + (4-|X|) = g(t) + 4-2|X|. So, as |X| \geq 2, we get g(t+1) \leq g(t).

Now, we have 4n = g(T) \leq g(0) \leq 4p. Thus, n \leq p, as desired.