The answer to the unparenthesized version is also negative. First, the topology induced by $d$ cannot be indiscrete, hence we can fix a $d$-closed set $F$ and points $a\in F$, $b\notin F$. For every Euclidean closed set $A$, let $f_c(x)=a+c\operatorname{dist}(x,A)$. Then $f$ is uniformly continuous, hence $d$-continuous, hence the $d$-closed set $f^{-1}(F)$ includes $A$ and excludes all points of Euclidean distance $b/c$ from $A$. The intersection of such sets for all $c$ is just $A$. This shows that the topology induced by $d$ refines the Euclidean topology.

If the topology is strictly finer, there exists a Euclidean convergent sequence $\{a_n\}\to a$ such that $a$ is not in the closure of $\{a_n\}$ in $(\mathbb R,d)$. By chosing a subsequence, we may assume that $a_n$ is strictly monotone, say increasing. Then for every strictly increasing sequence $\{b_n\}$ with supremum $b$, we can find an increasing continuous function $f\colon[b_0,b]\to[a_0,a]$ such that $f(b_n)=a_n$. We can extend $f$ to a uniformly continuous function $\mathbb R\to\mathbb R$. By assumption, $f$ is continuous under $d$, hence $\{b_n\}\not\to b$ in $d$. Since $-x$ is also $d$-continuous, the same holds for every decreasing sequence. This implies that every $d$-convergent sequence is eventually constant: any such sequence must converge in the Euclidean topology, and it if it were not constant, it would have a strictly monotone subsequence, which we already excluded. Thus, $d$ is discrete, hence every function $\mathbb R\to\mathbb R$ is uniformly continuous, a contradiction.

On the other hand, if $d$ induces the Euclidean topology, the property also clearly fails.

EDIT: The answer now applies to arbitrary topologies, using an idea by Pietro Majer from the comments.

Proposition: There are no topologies $\tau_0,\tau_1$ on $\mathbb R$ such that $f\colon\mathbb R\to\mathbb R$ is uniformly continuous in the Euclidean metric iff $f\colon(\mathbb R,\tau_0)\to(\mathbb R,\tau_1)$ is continuous.

Proof: $\tau_1$ cannot be indiscrete (lest all functions are uniformly continuous), hence we can fix a $\tau_1$-closed set $F$ and points $a\in F$, $b\notin F$. For every Euclidean closed set $A$ and $c>0$, let $f_c(x)=a+c\operatorname{dist}(x,A)$. Then $f_c$ is uniformly continuous, hence continuous from $(\mathbb R,\tau_0)$ to $(\mathbb R,\tau_1)$, hence the $\tau_0$-closed set $f_c^{-1}(F)$ includes $A$ and excludes all points of Euclidean distance $(b-a)/c$ from $A$. The intersection of such sets for all $c$ is just $A$. This shows that $A$ is $\tau_0$-closed, i.e., $\tau_0$ refines the Euclidean topology.

Let $f\colon\mathbb R\to\mathbb R$ be a Euclidean-continuous but not uniformly continuous function, such as $f(x)=x^2$. For every $n>0$, $f_n=f\restriction[-n,n]$ can be extended to a uniformly continuous function on $\mathbb R$. By assumption, this function is continuous from $(\mathbb R,\tau_0)$ to $(\mathbb R,\tau_1)$, hence $f_n$ is continuous from $([-n,n],\tau_0)$ to $(\mathbb R,\tau_1)$. Since $\tau_0$ refines the Euclidean topology, every point has a $\tau_0$-open neighbourhood included in some $[-n,n]$, thus $f=\bigcup_nf_n$ is continuous from $(\mathbb R,\tau_0)$ to $(\mathbb R,\tau_1)$. However, it is not uniformly continuous in the Euclidean metric, a contradiction.