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.)
The fact that InjRad is lower semicontinuous is obvious; i.e. if $x_n\to x$ then $$\liminf\ \mathrm{InjRad}_{x_n}\ge \mathrm{InjRad}_x$$ Now let me show that if it is true for compact manifolds then the same true for complete ones.
If $R<\mathrm{InjRad}_p$ then one can construct a smooth metric on a sphere with an isometric copy of $B_R(p)$ inside.
Now if
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..contradiction again.
