If $F: E \to E'$ is an isomorphism of vector bundles, how can I show the restriction to each fiber $F_p : E_p \to E'_p$ is a linear isomorphism?
Since $F$ is a isomorphism of vector bundles, it is a homeomorphism.
I'm having trouble showing $F_p$ is bijective. I know $F_p$ is linear, but I don't see a reason why this must automatically be bijective.
I tried using the local trivialization $E_p, E'_p \cong \{p\} \times \mathbb R^k$, but no luck.
To expand on my comment.
We can do this twice in two different contexts.
Your apparent context: It appears you are considering the category of (cts/smooth/etc) vector bundles over a fixed base. In this category bundle maps from $E$ to $E'$ are (cts/smooth/etc) maps $F:E\to E'$ such that $\pi' F = \pi$, where $\pi:E\to B$ and $\pi': E'\to B$ are the projections.
Then if $p\in B$, and $x\in E_p = \pi^{-1}(p)$, then $\pi'(F(x)) = \pi(x)=p$. Thus $F(x)\in E'_p$. Therefore, we can see that for each $p\in B$, we have a functor $E\mapsto E_p$ from vector bundles over $B$ to vector spaces. Since functors take isomorphisms to isomorphisms, an isomorphism $\phi:E\to E'$ induces an isomorphism between $E_p$ and $E'_p$.
Now we can generalize.
The general case: Consider the category consisting of pairs $(E,B,\pi,b)$ with $E$ a vector bundle over $B$, $\pi:E\to B$ the projection, and $b\in B$ a point.
Morphisms from $(E,B,\pi,b)$ to $(E',B',\pi',b')$ are bundle maps $(F,f)$ with $F:E\to E'$, $f:B\to B'$ such that $f(b)=b'$, and the diagram $$\require{AMScd} \begin{CD} E @>F>> E' \\ @V\pi VV @VV\pi' V\\ B @>f>> B' \end{CD} $$ commutes.
Now we have a functor which sends the object $(E,B,\pi,b)$ to $E_b := \pi^{-1}(b)$. Checking that a morphism $(F,f)$ sends $E_b$ to $E'_{b'}$ is the same as in the earlier case.