The set $\mathcal{J} \subset \mathbb{R}^2$ defined by $\mathcal{J} = \{(x,y) \in \mathbb{R}^2\ |\ x=y\}$ has no interior points So far, I've though of it this way:
Assume it does, so there exists $(a,b) \in int(\mathcal{J})$ st. $B_{\delta}((a,b)) \subseteq \mathcal{J}$. Since $(a,b) = \alpha(1,1)$, then given $(j,k) \in B_{\delta}((a,b))$, $\sqrt{(j-\alpha)^2+(k-\alpha)^2}<\delta$, for $(j,k) \in \mathcal{J}$, then $(j,k) = \beta(1,1)$, so $\sqrt{2}(\beta-\alpha)<\delta$. I'm lost here. Is it possible to do that? How can I prove this?
Thanks

This is not the way to go to get a contradiction since it is not the case that an open ball around a point of $\mathcal J$ does not contain any other point of $\mathcal J$. In fact, any open ball $B_\delta(k,k)$ contains infinitely many points of $\mathcal J$.
The goal is to show that given $(k,k)\in\mathcal J$, there is some point in $B_\delta(k,k)$ which is not in $\mathcal J$ to show that $B_\delta(k,k)\subsetneq\mathcal J$
To do this, just view it geometrically. A ball around a point on a straight line does have points that are on that line, but also others that are not on that line. The simplest way to do it algebraically is to perturb one component a bit (as in @Nightflight's answer)
A point $(x,y)\in B_\delta(k,k)$ satisfies $(x-k)^2+(y-k)^2\lt\delta^2$. Our goal is to write a point $(x',y')$ that is in this ball but does not have $x'=y'$.
For $\delta\gt 0$, choose your favorite $\delta_1$ such that $0\lt\delta_1\lt\dfrac{\delta}{\sqrt 2}$ and $\delta_2:=\sqrt{\delta^2-\delta_1^2}$
By construction, $\delta_1\ne \delta_2$ and $\delta_1^2+\delta_2^2\lt\delta^2$ and note that $(k+\delta_1,k+\delta_2)\in B_\delta(k,k)$ but $\notin\mathcal J$ since $k+\delta_1\ne k+\delta_2$
This is one of many ways to construct such a point. The crux is to just choose two $\delta_1,\delta_2\gt 0$ such that $\delta_1\ne\delta_2$ and $\delta_1^2+\delta_2^2\lt \delta^2$ for any given $\delta\gt 0$ and note that $(k+\delta_1,k+\delta_2)$ is a point that is in the $\delta$-ball around $(k,k)$ but not in $\mathcal J$
