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In $\mathbb{R}^n$ define $d(x, y) = \begin{cases}1 \text{ if } x\neq y\\ 0 \text{ if } x = y\end{cases}$ to be a metric. Does $d$ induce the usual topology on $\mathbb{R}^n$?

Now to me it seemed that this must be obviously false, as continuity for functions in $\mathbb{R}^n$ would be greatly affected. So I tried to construct a conterexample.

My Attempted Proof

Assume that $d$ does induce the usual topology on $\mathbb{R}^n$. The topology induced by $d$ has as its basis $\mathcal{B} = \left\{B_d(x, \epsilon) \ | \ x \in \mathbb{R^n}, \epsilon > 0 \right\}$, and where each $B_d(x, \epsilon) = \{\ y \ | \ d(x, y) = 1 < \epsilon\}$.

Let $\mathcal{B}'$ denote the basis for the standard topology on $\mathbb{R}^n $. Since $d$ induces the usual topology on $\mathbb{R}^n$ we have $\mathcal{B}' \subset \mathcal{B}$, so that $U \in \mathcal{B}' \implies U \in \mathcal{B}$.

Take $x = (x_1, ... , x_n) \in \mathbb{R}^n$. Put $\delta = \epsilon < 1$. Then there exists a $U \in \mathcal{B}'$ such that $U = (x_1 - \delta, x_1 +\delta) \ \times \ ... \ \times \ (x_n - \delta, x_n +\delta)$ . However $\not\exists \ B_d(x, \epsilon)$ such that $B_d(x, \epsilon) = U$ as $U$ is nonempty, but $B_d(x, \delta = \epsilon < 1) = \emptyset$.

So that $\mathcal{B}' \not\subset \mathcal{B}$ and a contradiction is reached. Therefore $d$ does not induce the standard topology on $\mathbb{R}^n$. $\square$

Is my proof correct? If so how rigorous is it? Also any comments on my proof writing and any criticism on my proof is greatly appreciated.

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    The sequence $1,\frac12,\frac13,\dots$ converges to $0$ in the usual topology of $\mathbb R.$ Does it converge to $0$ in the topology given by that funny metric? (Hint: what happens for $\varepsilon=\frac12?$) – 2017-01-21

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Your proof seems correct to me, however let me propose you an alternative.

First I assume that you mean

$$d(x,y)=\begin{cases} 1 &\text{ if $x\ne y$} \\0 &\text{ if $x= y$}\end{cases}$$

otherwise it is not even a distance.

Then you know that every singleton $\{x\}$ for $x\in \mathbb R^n$ is open since the open ball $B(x,\frac12)$ contain $\{x\}$ but no other point.

You know that in the usual topology $\{x\}$ is not open, so these are not the same topologies.

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    Your assumption is correct, I forgot to add the cases for the metric. Thanks for pointing that out, the OP has been edited to reflect that. – 2017-01-21
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    @Perturbative another simple proof is that, in the euclidean topology, there exist uncountable compact sets in $\mathbb{R}^n$, while in the topology induced by the metric $d$ only finite sets are compact. – 2017-01-21