Are the rational numbers Q, with the relative topology as a subset of the real numbers, locally compact ? why?
You wrote locally path connected, when I assume you meant compact like you said in the actualy body.
The answer is no. Let $\displaystyle U$ be any neighborhood of $\displaystyle 0$ in $\displaystyle \mathbb{Q}$. Then, $\displaystyle \text{cl}_\mathbb{Q}\text{ }U=\mathbb{Q}\cap\text{cl}_\mathbb{R}\text{ }U$. Also, remember then that if $\displaystyle X,Y,Z$ are top. spaces and $\displaystyle X$ is a superspace of $\displaystyle Y$ then $\displaystyle Z$ will be a compact subspace of $\displaystyle Y$ if and only if it's a compact subspace of $\displaystyle X$ and thus for $\displaystyle \text{cl}_\mathbb{Q}\text{ }U$ to be compact in $\displaystyle \mathbb{Q}$ we need that $\displaystyle \mathbb{Q}\cap\text{cl}_\mathbb{R}\text{ }U$ is compact in $\displaystyle \mathbb{R}$. But, evidently this is not compact in $\displaystyle \mathbb{R}$, why?