Note that $t\mapsto \log t$ is uniformly continuous on $[1,\infty)$ (proven below). Note also that $t\mapsto 1+\sqrt{1+t^2}$ is uniformly continuous on $\mathbb R$ (also proven below). As the composition of uniformly continuous functions is uniformly continuous, the result follows.
To see that $\log$ is uniformly continuous on $[1,\infty)$, fix $\varepsilon>0$. Assume that $1\leq x, $y-x<\delta$. Then $y/x<1+\delta/x\leq1+\delta$, and so $ \log y-\log x=\log \frac y x\leq\log(1+\delta); $ if we choose $\delta$ small enough so that $\log(1+\delta)<\varepsilon$, we are done.
For the uniform continuity of $g:t\mapsto 1+\sqrt{1+t^2}$, fix $\varepsilon>0$. Choose $x_0$ such that $\sqrt{1+x^2}>3/\varepsilon$ if $|x|\geq x_0$. Since $g$ is continuous on the compact set $[-x_0-1,x_0+1]$, it is uniformly continuous there. So there exists $\delta_1>0$ such that $x,y\in[-x_0,y_0]$ with $|y-x|<\delta_1$ implies $|g(y)-g(x)|<\varepsilon$.
Now let $\delta=\min\{\delta_1,\varepsilon/3,1\}$. Suppose that $|x-y|<\delta$. If both $|x|, then $|y| and so $|g(y)-g(x)|<\varepsilon$ by the uniform continuity on the compact set. If $|x|\geq|x_0$, then $ |g(y)-g(x)|=|\sqrt{1+y^2}-\sqrt{1+x^2}|\leq|\sqrt{1+y^2}-|y||+||y|-|x||+||x|-\sqrt{1+x^2}|\\ \leq\frac1{|y|+\sqrt{1+y^2}}+|y-x|+\frac1{|x|+\sqrt{1+x^2}}<\frac1{\sqrt{1+y^2}}+\frac1{\sqrt{1+x^2}}+\frac\varepsilon3\\ <\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon. $