Suppose that $V$ is a vector space, and let $V → W$ be a linear map.
$$V_0 ⊆ V_1 ⊆ · · · ⊆ V_i ⊆ V_{i+1} ⊆ · · · ⊆ V$$
are subspaces of $V$ (one for each $i = 0, 1, 2, \ldots$) and inclusions thereof, such that $V$ is the union of all $V_i$.
Show that $f$ is injective if and only if the restriction of $f$ to every $V_i$ is injective. (The restriction $f$ to $V_i$ is the composition $V_i\hookrightarrow V\stackrel f\to W$, where the first map it the inclusion.)
One direction is clear: If $f$ is injective then so is each $f|_{V_i}$.
For the other direction assume all restrictions are injective and let $v,w\in V$ with $f(v)=f(w)$. Then $v\in V_i$ and $w\in V_j$ for some $i,j\in\Bbb N$. With $k=\max\{i,j\}$ we have $v,w\in V_k$ and by injectivity of $f|_{V_k}$ we get $v=w$ as desired.