= Solution
A <sequence> $f_n:S\to\mathbb R$ converges uniformly to $f$ if, for every $\varepsilon>0$, there is $N$ such that
$$
|f_n(x)-f(x)|<\varepsilon
$$
for every $x\in S$ and every $n\geq N$. A map $h:M\to N$ is uniformly continuous if, for every $\varepsilon>0$, there is $\delta>0$ such that
$$
d_M(x,y)<\delta\quad\Longrightarrow\quad d_N(h(x),h(y))<\varepsilon
$$
for all $x,y\in M$.
If $f\in C_0(\mathbb R^d)$, choose a ball outside which $|f|<1$. On the closed ball, continuity gives boundedness, so $f$ is bounded everywhere.
Now let $(f_n)$ be Cauchy in the uniform metric. For each $x$, $(f_n(x))$ is Cauchy in $\mathbb R$; let its <limit> be $f(x)$. Passing to the pointwise <limit> in the uniform Cauchy estimate shows that $f_n\to f$ uniformly. Hence $f$ is continuous. Given $\varepsilon>0$, choose $n$ with $\|f-f_n\|_\infty<\varepsilon/2$, and then choose $K$ so that $|f_n(x)|<\varepsilon/2$ for $\|x\|>K$. It follows that $f$ also vanishes at infinity. Thus $C_0(\mathbb R^d)$ is complete, as in <completeness of continuous functions vanishing at infinity>.
Every $f\in C_0(\mathbb R^d)$ is uniformly continuous. Given $\varepsilon>0$, choose $R$ so that $|f(x)|<\varepsilon/2$ outside the ball of radius $R$. On the compact ball of radius $R+1$, $f$ is uniformly continuous; choose the corresponding $\delta\leq1$. If two points at distance below $\delta$ are not both in that ball, then both lie outside the ball of radius $R$, and their <function> values differ by less than $\varepsilon$.
For the final <sequence>, continuity of $\varepsilon$ at zero gives, for each fixed $x$,
$$
f_n(x)=\sqrt{x^2+\varepsilon(x/n)}\longrightarrow|x|.
$$
Thus <pointwise convergence> is compulsory. <Uniform convergence> need not hold: with $\varepsilon(t)=t^2$,
$$
f_n(x)-|x|
=|x|\left(\sqrt{1+n^{-2}}-1\right),
$$
which is unbounded as a <function> of $x$ for every fixed $n$.
Under the additional bound $\varepsilon(t)\leq M|t|$, however,
$$
0\leq f_n(x)-|x|
=\frac{\varepsilon(x/n)}{\sqrt{x^2+\varepsilon(x/n)}+|x|}
\leq\frac Mn
$$
for $x\ne0$, and the difference is zero at $x=0$. Hence convergence is uniform, by the <uniform square-root perturbation under linear growth> estimate. The pointwise answer remains yes.
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