Prove Addition is Continuous (without epsilon-delta!)

How about using Sequential criteria?

Let $$(x_n,y_n) \to (x,y)$$ Then $$x_n \to x, y_n \to y$$ Thus, $$x_n+y_n \to x+y$$ Hence, addition is continuous.

I think if you want to avoid the $\varepsilon - \delta$ discussion then you should use the "preimage of open is open" definition of discontinuity. I just saw Mathmoore's answer so I guess my answer is really a rephrasing of his.

The topology on $\mathbb{R}$ is generated by open intervals. Hence it suffices to show that the preimage under addition of an open interval $(a, b)$ is an open subset of $\mathbb{R}^2$. The preimage is of course the set $\{(x, y)\in\mathbb{R}^2: a < x + y < b\}$. The solutions to the inequality $x + y < b$ is an open halfspace with base line given by the equation $x + y = b$. Similarly, the solutions to the inequality $a < x + y$ is an open subspace where the base line is given by the equation $x + y = a$ (paralel to the previous line!). Using that "finite intersections of open is open", or more simply that the intersection of the two open halfspaces is the open strip between the lines $x + y = a$ and $x + y = b$, we get that "$+$" is continuous (with respect to the standard topologies on $\mathbb{R}$ and $\mathbb{R}^2$.

If $I$ and $J$ are open in $\mathbb R$ then $I\times J$ is open in $\mathbb R^2.$ To prove continuity of $f(x,y)=x+y,$ it suffices to show that when $f(x,y)\in K$, where $K$ is open in $\mathbb R,$ there exist open real $I,J$ such that $(x,y)\in I\times J$ and $f(I\times J)\subset K.$

There exists $r>0$ such that $(-r+f(x,y),r+f(x,y))\subset K.$ So let $I=(-r/2+x,r/2+x)$ and $J=(-r/2+y,r/2+y).$

We cannot avoid mentioning $r$. With a different topology on $\mathbb R,$ addition may fail to be continuous.

In general, $f:X\to Y$ is continuous iff whenever $f(p)\in V$ with $V$ open in $Y,$ there exists $U,$ open in $X$, with $p\in U$ and $f(U)\subset V.$ This could be called the topological generalization of the classical "$\epsilon$-$\delta$" definition of continuity.