You have the right idea: it’s true that no open ball centred at your point $y$ is contained in $U(0,1)$. To finish the argument, you have to show this, and then you have to generalize the argument to all sets of the form $U(x,r)$.
Consider the ball $B_{\bar\rho}(y,r)$, where $r>0$; we want to show that it’s not a subset of $U(0,1)$ no matter how small $r$ is. Clearly $y$ is nudging the ‘top edge’ of $U(0,1)$, so we should try to pick something ‘bigger than’ $y$, but still ‘small’ enough to fit in $B_{\bar\rho}(y,r)$; since $$y=\left(\frac{n-1}{n}\right)_{n\in\mathbb{Z}^+}=\left(1-\frac1n\right)_{n\in\mathbb{Z}^+},$$ a natural idea is $$z=\left(\frac{n-1}{n}+\frac{r}2\right)_{n\in\mathbb{Z}^+} = \left(1-\frac1n+\frac{r}2\right)_{n\in\mathbb{Z}^+}.$$ Then $\bar\rho(y,z) = \sup\limits_{k\in\mathbb{Z}^+}\min\{|y_k-z_k|,1\}=\min\{r/2,1\}\le r/2$, so $z\in B_{\bar\rho}(y,r)$. But if $n\ge \dfrac2r$, then $d\frac1n \le \dfrac{r}2$, $z_n = 1 - \dfrac1n + \dfrac{r}2 \ge 1$, and hence $z\notin U(0,1)$.
This shows that $U(0,1)$ is not open, and the same idea generalizes very easily. To show that an arbitrary $U(x,\epsilon)$ is not open, just scale this example by a factor of $\epsilon$ and translate it from the origin to $x$. (This is what Davide was suggesting in his comment.) That is, instead of taking $$y=\left(\frac{n-1}{n}\right)_{n\in\mathbb{Z}^+},$$ let $$y=x+\left(\frac{n-1}{n}\epsilon\right)_{n\in\mathbb{Z}^+}=\left(x_n+\frac{n-1}{n}\epsilon\right)_{n\in\mathbb{Z}^+}.$$ Then $y\in U(x,\epsilon)$, but no ball centred at $y$ is a subset of $U(x,\epsilon)$: the argument used above works with only very minor modifications, which I’ll leave to you.
Best Answer
Use the equivalent definition of local compactness: if $x \in U \subset C$, $U$ open, $C$ compact, then there must be a ball $B_{\epsilon}(x)$ whose closure $\bar{B}$ is contained in $U$. Apply this to $x = 0$. Then the closure of such a ball is compact as it is a closed subset of a compact set in a metric space. But it is not: look at the set $A = \{ x_i = (0, \dots, 0, \epsilon, 0, \dots ), i \in \mathbb{N} \} \subset \bar{B}$, with the $\epsilon$ in position $i$. It has no limit point $x$ in $A$ ($x$ cannot contain a coordinate in $(0, \epsilon)$ or a small enough ball around $x$ will not contain any $y \in A$; and in the other case, it is at distance $0$ or $\epsilon$ from any point in $A$ - think about that!). So $A$ is not limit-point compact, which in metric spaces is equivalent to being compact. Contradiction.