Just wanted to verify if the following proof works:
Suppose $f:X\rightarrow \mathbb{R}$ $g:X\rightarrow \mathbb{R}$ are continuous.
Want to show that: $h=f+g$ is continuous.
We are going to prove the statement using the localised definition of continuity: https://www.emathzone.com/tutorials/general-topology/continuity-in-topological-spaces.html
Let $x\in X$ and $U$ is open in $\mathbb{R}$ s.t $h(x)\in U$. Hence, $\exists \epsilon>0$ s.t $(h(x)-\epsilon,h(x)+\epsilon)\subset U$.
Since $f$ and $g$ are continuous at $x$ and $$f(x)\in B(f(x),\epsilon/2)$$ and $$g(x)\in B(g(x),\epsilon/2)$$ which are open sets in $\mathbb{R}$, this implies the $\exists V_1, V_2$ open in $X$ s.t $x\in V_1\cap V_2$ and $$f(V_1)\subset B(f(x),\epsilon/2) $$ and
$$g(V_2)\subset B(g(x),\epsilon/2)$$
Hence, $\forall z\in V_1\cap V_2$ $$|h(z)-h(x)|\leq|f(z)-f(x)|+|g(x)-g(z)|< \epsilon$$
i.e $h(V_1\cap V_2)\subset(h(x)-\epsilon,h(x)+\epsilon)\subset U$.
Since $V_1\cap V_2$ is open in $X$ and $x\in V_1\cap V_2$, this implies $h$ is continuous at $x$.
Best Answer
An alternative proof, if you know that $p: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ defined by $p(x,y)=x+y$ is continuous already:
Define $f \nabla g: X \to \mathbb R \times \mathbb R$ by $(f\nabla g)(x)=(f(x), g(x))$ and note that $f \nabla g$ is continuous as $\pi_1 \circ (f \nabla g) = f$ and $\pi_2 \circ (f \nabla g) = g$ are both continuous and $\mathbb{R}\times \mathbb{R}$ carries the product topology.
Then $f+g=h=p\circ (f \nabla g)$ is continuous as the continuous composition of continuous maps.
But your pointwise proof is fine too.