$\newcommand{\ts}{\tilde s}$Yes, this is true.
Indeed, take any real $c\ge1/2$ and let \begin{equation} s_n(y):=\sum_{-n^2\le k\le n^2}\frac1{(ny-k)^2} \end{equation} and \begin{equation} \ts_n(x):=\inf_{|y-x|<c/n}s_n(y). \end{equation}
Proposition 1: $\ts_n(x)\to\pi^2$ (as $n\to\infty$) uniformly over all real $x$ such that $n^2-n|x|\to\infty$.
Proof: Suppose that $n$ is large enough so that $n^2-n|x|>c+1$. Take any real $y$ such that $|y-x|<c/n$, so that $n|y|<n^2-1$. If $n|y|$ is an integer, then $s_n(y)=\infty$, so that excluding such values of $y$ does not affect the value of $\ts_n(x)$. Now, excluding such values of $y$ indeed and letting \begin{equation} m:=\lfloor ny\rfloor, \end{equation} we get \begin{equation} s_n(y)=s_{n,m}(u):=\sum_{-n^2+m\le j\le n^2+m}\frac1{(u+j+1/2)^2}, \end{equation} where \begin{equation} u:=ny-(m+1/2)\in(-1/2,1/2). \end{equation}
For $q=1,2,\dots,\infty$, let \begin{equation} t_q(u):=\sum_{-q-1< j<q}\frac1{(u+j+1/2)^2}. \end{equation} Then $t_q$ is an even convex function on the interval $(-1/2,1/2)$, so that \begin{equation} t_q\ge t_q(0) \end{equation} and, moreover, \begin{equation} t_\infty(0)=\pi^2. \end{equation} Furthermore, by the convexity of $\frac1{z^2}$ in $z>0$, \begin{equation} \pi^2-t_q(0)=t_\infty(0)-t_q(0)=2\sum_{j\ge q}\frac1{(j+1/2)^2} <2\int_q^\infty\frac{dz}{z^2}=\frac2q. \end{equation}
So, for $y$ as described above, \begin{equation} s_n(y)=s_{n,m}(u)\ge t_{n^2-|m|}(u)\ge t_{n^2-|m|}(0) >\pi^2-\frac2{n^2-|m|}\to\pi^2, \end{equation} since $n^2-n|x|\to\infty$, $|m|\le n|y|+1$, and $|y-x|<c/n$, so that $n^2-|m|\ge n^2-n|x|-(c+1)\to\infty$, so that \begin{equation} \ts_n(x)=\inf_{|y-x|<c/n}s_n(y)\ge \pi^2-o(1). \end{equation}
On the other hand, letting $l:=\lfloor nx\rfloor$, we have $|\frac{l+1/2}n-x|\le\frac1{2n}\le\frac cn$, so that \begin{equation} \ts_n(x)=\inf_{|y-x|<c/n}s_n(y)\le s_n\Big(\frac{l+1/2}n\Big) =s_{n,l}(0)<t_\infty(0)=\pi^2. \end{equation}
Proposition 1 follows immediately from the latter two displays. $\quad\Box$