The compactness is irrelevant; i.e if it is true for compact manifolds then the same is true for complete ones. (The same proof as in comact case works, but it is easier to do this way.)

If $R<\mathrm{InjRad}_p$ then one can construct a smooth metric on a sphere with an isometric copy of $B_R(p)$ inside.

If there is a sequence of points $x_n\to x$  such that
$$\lim\ \mathrm{InjRad}_{x_n}< \mathrm{InjRad}_x,$$
apply above consruction for $R$ slightly smaller than $\mathrm{InjRad}_x$.
You get a compact manifold with non-continuous InjRad.

If there is a sequence of points $x_n\to x$ such that 
$$\lim\ \mathrm{InjRad}_{x_n}> \mathrm{InjRad}_x$$ then apply above construction for $p=x_n$ for large enough $n$ and $R>\mathrm{InjRad}_x$.
That leads to a contradiction again.