I think that Stephen's idea can be adapted to show that the injectivity radius is also upper semicontinuous in the incomplete case. Here's my argument -- let me know if you see anything wrong.

Let $(M,g)$ be a connected Riemannian $n$-manifold, and let $i\colon M\to(0,\infty]$ be the injectivity radius function. Suppose $i$ is not upper semicontinuous at $x\in M$, and let $r = i(x)$. Then there exists a sequence of points $x_k\to x$ and a number $R>r$ such that $i(x_k)\ge R$ for all $k$. Let $\varepsilon = (R-r)/3$, $R'=r+\varepsilon$, and $R'' =r+2\varepsilon$, so that $r<R'<R''<R$. Choose $k_0$ large enough that $d(x_k,x)$ and $d(x_k,x_{k_0})$ are both less than $\varepsilon$ for all $k\ge k_0$. Since the geodesic balls $B_r(x)$, $B_{R'}(x_k)$, and $B_{R''}(x_k)$ are also metric balls, the triangle inequality implies that for each $k\ge k_0$,
$$
B_r(x) \subseteq B_{R'}(x_k) \subseteq B_{R''}(x_{k_0}).
$$

Using normal coordinates, we can identify $B_R(x_{k_0})$ with the Euclidean ball $B_R(0)\subset \mathbb R^n$, and then by using a bump function we can create a complete metric $\hat g$ on $\mathbb R^n$ that agrees with $g$ on $\overline B_{R''}(x_{k_0})$. The set inclusions above imply that $g$ and $\hat g$ agree on $B_{R'}(x_k)$ for each $k$, and thus $\hat i(x_k)\ge R'$ for all $k\ge k_0$. By continuity of $\hat i$, we have $\hat i(x) = \lim_k \hat i(x_k) \ge R'$.

This means that the exponential map of $\hat g$ is injective on the ball of radius $R'$ in $T_xM$. If we can show that the exponential map of $\hat g$ is equal to that of $g$ on that ball, then we have a contradiction to the assumption $i(x)=r<R'$.

Let $\gamma\colon [0,R')\to \mathbb R^n$ be a unit-speed $\hat g$-geodesic starting at $x$, and let
$$t_0 = \sup\{t\in [0,R'): \gamma(t) \in B_{R''}(x_{k_0})\}.$$
Then for all $0\le t < t_0$, $\gamma(t)$ is in the set where $g=\hat g$, and thus $\gamma|_{[0,t_0)}$ is also a $g$-geodesic. We need to show that $t_0=R'$ for every such $\gamma$.

Suppose $t_0<R'$ for some such $\gamma$. Then $\gamma(t_0)\in \partial B_{R''}(x_{k_0})$, which means that $d_g(x_{k_0},\gamma(t_0)) = R''$. However,
\begin{align*}
d_g(x_{k_0},\gamma(t_0))
&\le d_g(x_{k_0},x) + d_g(x,\gamma(t_0))\\
&< \varepsilon + L_g(\gamma|_{[0,t_0]})\\
&= \varepsilon + t_0\\
&< \varepsilon + R' \\
&= R'',
\end{align*}
which is a contradiction.