Suppose that $V$ is a vector space such that for every net $(v_i)_{i \in I}$ converging to $v$ and for every net $(w_i)_{i \in I}$ converging to $w$, we have that $(v_i + w_i)_{i \in I}$ converges tot $v+w$. Show that $+: V \times V \to V$ is a continuous map.
Attempt:
We simply show net continuity, which is equivalent with the topological continuity we want. That is, assume $g: J \to V \times V$ is a net in $V \times V$ converging to $(v,w)$. Our goal is to show that the net $+ \circ g: J \to V$ converges to $v+w$.
Since $V\times V$ carries the product topology, the projections $\pi_1, \pi_2$ are continuous and hence $\pi_1 \circ g: J \to V$ converges to $x$ and $\pi_2 \circ g: J \to V$ converges to $y$.
Consequently, our hypothesis implies that $\pi_1 \circ g + \pi_2 \circ g = + \circ g: J \to V$ converges to $v+w$, as desired.
Is this correct? (Sorry for writing my nets cryptic like this, but it is just to make sure everything is clear).
Suppose that $+$ were not continuous at $(v_0,w_0)$ so that there is some open neighbourhood $O$ of $v_0+w_0$ such that for every neighbourhood $U$ of $(v_0,w_0)$ there is some $(v',w') \in U$ with $v'+w' \notin O$.
Now, let $J$ be the set $\{U \times V: U,V \text{ open in } V, v_0 \in U, w_0 \in V\}$ ordered by reverse inclusion. For every $j=U \times V$ in $J$ we pick $n(j):=(v',w') \in U \times V$, such that $\pi_1(n(j))+\pi_2(n(j)) \notin O$, by our non-continuity assumption above.
It's clear by definition of the product topology that $\pi_1 \circ n: J \to V \times V$ converges to $v_0$ and likewise $\pi_2 \circ n$ converges to $w_0$ while nevertheless $+ \circ ((\pi_1 \circ n) \times (\pi_2 \circ n))$ does not converge to $v_0 + w_0$ contrary to your assumption.