Let $E_{a,b}$ be an elliptic curve defined by the equation $y^{2}=x^3+ax+b$ where $a,b \in \mathbb{Q}$. Suppose that for $a=a_{0}$ and $b=b_{0}$ the rank of $E_{a_{0},b_{0}}(\mathbb{Q})=1$.
question: is there an $\epsilon> 0$ such that for any $(\alpha,\beta) \in \mathbb{Q}^{2}$ and $|\alpha-a_{0}|+|\beta -b_{0}|< \epsilon$ then the rank of $E_{\alpha,\beta}(\mathbb{Q})=1$.
A counter-example, showing that the answer is "no" for some $(a_0,b_0)$ can be constructed as follows. Take an elliptic surface over $\mathbb{Q}$ with rank $2$. By Silverman's specialisation theorem, the rank of each fibre is at least $2$ (but often larger) except for a finite number of fibres. At an exceptional fibre of rank $1$, you will get a counter-example.
For instance $a_0=2$, $b_0=1$ the curve is of rank $1$, but the family $E_t:y^2=x^3+(−t^2+t+2)x+1$ is of rank at least $2$ over $\mathbb{Q}(t)$ as it contains $(0,1)$ and $(t,t+1)$. For all $t$ close to zero the rank of $E_t(\mathbb{Q})$ will be $2$ or larger.
I believe a similar argument should work in general and show that the answer is always "no".
Arithmetic properties like the rank of the Mordell-Weil group are not continuous in the real or $p$-adic topology. For instance, the number of prime factors of numerator and denominator of $\Delta$ will have an influence on the rank.
In the above example as $t=1/n$ approaches $0$, the rank will jump around $2$ and $3$ and sometimes larger values rather randomly: $$ \begin{array}{c|cccc} t=\tfrac{1}{n} & \tfrac{1}{2} & \tfrac{1}{3} & \tfrac{1}{4} &\tfrac{1}{5} & \tfrac{1}{6} & \tfrac{1}{7} & \tfrac{1}{8} & \tfrac{1}{9} & \tfrac{1}{10} & \tfrac{1}{11} & \tfrac{1}{12} & \tfrac{1}{13} & \tfrac{1}{14} &\tfrac{1}{15} & \tfrac{1}{16} & \tfrac{1}{16} & \tfrac{1}{17} & \tfrac{1}{18} & \tfrac{1}{19}\\ \text{rank} & 1 & 3 & 2 & 2 & 3 & 2 & 3 & 3 & 3 & 3 & 4 & 2 & 2 & 3 & 3 & 3 & 3 & 3 &2 \\ \end{array} $$
