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.
Your intuition is right for the first question. Let $y_n = x_n + (1 - 2^{-n})\epsilon$; then clearly $y = (y_n) \in U(x,\epsilon)$. But no ball around $y$ in the uniform metric is contained in $U(x,\epsilon)$: if the radius of such a ball is $\delta$ and $\delta > 2^{-n}\epsilon$ for some $n$, then $B_\delta(y)$ contains, for example, $z = (z_k)$ where
$$z_k = \begin{cases} y_k & k \ne n \\ y_n + 2^{-n}\epsilon & k = n, \end{cases}$$
which is not in $U(x,\epsilon)$.
The second question is asking you to check two things. The first is that every $U(x,\delta)$ for $\delta < \epsilon$ is contained in $B_\epsilon(x)$. That is, every point $y$ in $U(x,\delta)$ has $\rho(x,y) < \epsilon$. But in fact, $\rho(x,y) \le \delta$, since $|y_n - x_n| < \delta$ for each $n$. (Make sure you understand why this is $\le \delta$ and not $< \delta$ -- this is the same principle behind my answer to the first question.) The second thing to check is that every $y$ with $\rho(x,y) < \epsilon$ is contained in some $U(x,\delta)$. If $\rho(x,y) = \epsilon' < \delta < \epsilon$, then you can easily check that $y$ is contained in $U(x,\delta)$.
Here's some 'philosophy' that may be helpful. Membership in these box-topology-style open sets like $U(x,\epsilon)$ is determined by each coordinate individually, but membership in uniform open sets is determined by the behavior of the entire sequence $(y_n)$, since you're taking a supremum.
Best Answer
Note that under the uniform topology the corresponding metric is $d(x,y) = \min(1, \sup_i (|x_i-y_i|))$. In a finite product of intervals $(x_i-\epsilon, x_i + \epsilon)$ any particular element will thus be strictly less than $\epsilon$ distance away from $x$, but in an infinite product you have elements like $y = (y_i) = (x_i + (1-1/n)\epsilon)$, where the supremum is equal to $\epsilon$. Note that any open neighborhood of this particular point $y$ will contain elements not in the first given product, demonstrating that the product is not open.
Such elements must be excluded from the basis elements, which should instead be the set of elements of distance strictly less than $\epsilon$ from $x$. This set can be constructed as the given union over $\delta <\epsilon$.