Spot the Error in the Proof that the Set of All Sections in a Family of Vector Spaces is itself a Vector Space A family of vector spaces $E$ over $X$ is a notion one starts with before defining a vector bundle. 
Concretely, the data is $(E,X)\in obj(Top)^2$, $p\in hom_{Top}(E,X)$, such that $\forall x \in X$, $p^{-1}(\{x\})\in obj(Vect_{\mathbb{F}}^n)$ (that is, it is endowed with the structure of a finite-dimensional vector space), such that local vector addition $a_x:p^{-1}(\{x\})^2\to p^{-1}(\{x\})$ and local scalar multiplication $m_x:\mathbb{C}\times p^{-1}(\{x\})\to p^{-1}(\{x\})$ are continuous with respect to the subspace topology on $p^{-1}(\{x\})$ which is inherited from the topology of $E$.
A section is a map $s\in hom_{Top}(X,E)$ such that $p\circ s= id_X$.
We may also define $\Gamma(E)$ as the set of all possible sections on a family $E$ of vector spaces over $X$ (note: we are specifically not assuming the local triviality condition which would have made $E$ a vector bundle).
Claim: $\Gamma(E)\in obj(Vect_{\mathbb{F}})$
Proof: We define vector addition in $\Gamma(E)$ as follows. If $(s_1, s_2)\in \Gamma(E)^2$ and $x\in X$ are given, then $(s_1(x), s_2(x))\in p^{-1}(\{x\})$, so that both $s_1(x)$ and  $s_2(x)$ are two points in the same vector space $p^{-1}(\{x\})$ and can thus be added using $a_x$. So define $s_1+s_2$ at this $x$ to be $a_x(s_1(x),s_2(x))$. If $i_x:p^{-1}(\{x\})\to E$ is the local inclusion map and $\eta:X\to X^2$ is given by $x\mapsto(x,x)$ then we may write $$ s_1+s_2 :=i_x \circ a_x\circ(s_1\times s_2)\circ\eta $$
Because $s_1+s_2$ is the composition of continuous maps, it is itself continuous. It also obeys the condition $p\circ(s1+s2)=id_X$ and thus it is also a section.
Similarly we define scalar multiplication as follows: If $\lambda\in\mathbb{F}$, then $$\lambda s := m_x\circ M_\lambda\circ s$$where $M_\lambda:E\to\mathbb{F}\times E$ is the map defined by $e\mapsto(\lambda, e)$. Again, this is the composition of continuous maps and it is thus continuous.

My question is: what is the fault in this proof? I know it is wrong because you need local triviality for this thing to be a vector space. I would furthermore appreciate if someone could also provide the actual proof given the $E$ is a vector bundle, and not merely point out where the fault in the provided proof is (though the latter is the main motivation for my question).
 A: The additions on the fibres give a global map
$$a \colon E \times_X E \to E,\quad a(\xi,\upsilon) = a_{p(\xi)}(\xi,\upsilon),$$
where $E \times_X E = \{(\xi,\upsilon) \in E^2 : p(\xi) = p(\upsilon)\}$ is the fibre product of $E$ with itself. In the same vein, we have a global multiplication
$$m\colon \mathbb{F}\times E \to E,\quad m(\lambda,\xi) = m_{p(\xi)}(\lambda,\xi)$$
If these maps are continuous, then the sum of two sections and scalar multiples of sections are again continuous, hence section, since we can write them as
$$s_1 + s_2 = a \circ (s_1,s_2)\quad \text{resp.} \quad m \circ M_\lambda \circ s,\tag{$\ast$}$$
and $(s_1,s_2) \colon X \to E \times_X E,\; x \mapsto (s_1(x),s_2(x))$ as well as $M_\lambda \circ s \colon X \to \mathbb{F}\times E,\; x \mapsto (\lambda, s(x))$ are continuous when $s,s_1,s_2$ are continuous.
If we have local triviality, then the global addition and multiplication are continuous, since a local trivialisation is a homeomorphism $h\colon p^{-1}(U) \to U \times \mathbb{F}^k$ that preserves fibres (that is, $\pi_1(h(e)) = p(e)$ for all $e\in p^{-1}(U)$) and is linear on each fibre. Such a local trivialisation induces a homeomorphism $\tilde{h}\colon \tilde{p}^{-1}(U) \to U \times \mathbb{F}^k \times \mathbb{F}^k$ which is a local trivialisation of the fibre product. The fibrewise addition $U\times \mathbb{F}^k \times \mathbb{F}^k \to U \times \mathbb{F}^k$ is continuous, and hence the global addition $a$ is continuous on the open subset $\tilde{p}^{-1}(U)$ of $E\times_X E$ (where $\tilde{p}$ is the projection of the fibre product to $X$). Since continuity is a local property, it follows that $a$ is then globally continuous. Similarly, but a little easier, it follows that the global scalar multiplication $m$ is continuous when we have local triviality.
And when we have local triviality, or just the continuity of $a$ and $m$, then $(\ast)$ immediately shows the continuity of the sum resp. scalar multiples of sections, hence that $\Gamma(E)$ is then a vector space.
Without assuming either directly the continuity of $a$ and $m$ or indirectly by requiring something that implies it, we cannot deduce that the "discontinuous sections" $s_1 + s_2$ and $\lambda s$ are in fact continuous.
As an example, let $X = \mathbb{R}$, and $\Phi \colon \mathbb{F}^k \to \mathbb{F}^k$ any homeomorphism. Then let $E = X \times \mathbb{F}^k$ and $p \colon E \to X$ the projection to the first factor. But for the vector space structure on the fibres, we do something non-obvious. For $x \neq 0$, we take the usual vector space operations on $\mathbb{F}^k$, but on $p^{-1}(\{0\})$, we define
$$a_0((0,\xi),(0,\upsilon)) = (0, \Phi^{-1}(\Phi(\xi) + \Phi(\upsilon))) \quad \text{and} \quad m_0(\lambda,(0,\xi)) = (0,\Phi^{-1}(s\cdot \Phi(\xi))).$$
Then the sum of two sections, and scalar multiples of sections generally are discontinuous at $0$. If we choose for example a translation, $\Phi(\xi) = \xi + \eta$ for some $\eta \neq 0$, then for the section $z \colon x \mapsto (x,0)$, we have
$$\pi_2((z+z)(x)) = \begin{cases} 0 &, x \neq 0 \\ \eta &, x = 0,\end{cases}$$
so $z+z$ is not continuous at $0$. And if $s$ is any section, then
$$\pi_2((0\cdot s)(x)) = \begin{cases} 0 &, x \neq 0 \\ -\eta &, x = 0.\end{cases}$$
In a few words, in this example the vector space structures on the fibres don't fit together. Local triviality ensures that the vector space structures on the fibres fit together.
