Let $f,h:E \to F$ be continuous functions between topological vector spaces. Then $f+h$ is continuous I'm trying to prove this result

Let $f,h:E \to F$ be continuous functions between topological vector spaces. Then $f+h$ is continuous.

I posted my proof as below answer. Could you have a check on my attempt?
 A: The proof of your lemma is overly complicated. The lemma follows immediately from continuity of the addition.
Actually, you don't even need the lemma to prove the claim. Simply note that $f+g$ can be seen as the composition $+_Y \circ (f,g)$ where $(f,g): X \to Y\times Y: x \mapsto (f(x), g(x))$ is continuous by elementary topology.
Note that nowhere it was used that $X$ is a topological vector space, so $X$ can in fact be any topological space and the result is still true. This shouldn't surprise us: the statement you mention never uses the vector space structure of $X$, but only the vector space structure on $Y$.
A: First, we need the following lemma.

Lemma: Let $E$ be a topological vector space and $(x_d)_{d\in D}, (y_d)_{d\in D}$ nets in $E$. Suppose that $x_d \to x$ and $y_d\to y$. Then $x_d+y_d \to x+y$.

Proof of the lemma: Consider the map $T: E \times E \to E, (a,b) \to a+b$. Then $T$ is continuous. Let $U$ be a neighborhood (nbh) of $x+y$. Then there is a nbh $V$ of $(x,y)$ such that $T[V] \subseteq U$. By construction of product topology, there is nbh $V_1$ of $x$ and a nbh $V_2$ of $y$ such that $V_1 \times V_2 \subseteq V$. By net convergence, there is $d_1,d_2 \in D$ such that $x_d \in V_1$ for all $d \ge d_1$ and $y_d \in V_2$ for all $d \ge d_2$. Let $d_3 := \max\{d_1, d_2\}$. Then $x_d+y_d \in U$ for all $d \ge d_3$. This completes the proof of the lemma.
Let $(x_d)$ be a net in $E$ such that $x_d \to x\in E$. By continuity of $f,h$, we get $f(x_d) \to f(x)$ and $h(x_d) \to h(x)$. It follows from our lemma that $$(f+h) (x_d) = f(x_d)+h(x_d) \to f(x)+h(x) = (f+h) (x).$$ Hence $f+h$ is continuous.
