Denote $D = \{ x\in \mathbb{R}^{d+1}: \sqrt{x_1^2+\dots+x_d^2 }\leq x_{d+1}\}$ which is the second order cone ( ice-cream cone) in $\mathbb{R}^{d+1}$. My question is that if a standard brownian motion $W_t=(X^1_t,\dots, X^{d+1}_t)$ in $\mathbb{R}^{d+1}$ is started at $x_0=(1,0,\dots,0)$, will it finally hit $D$, i.e., $$P^{x_0}( \exists t>0, W_t \in D)=1.$$
In the case $d=1$, it is true as $\{W_t:t>0\}$ is dense in $\mathbb{R}^2$. For $d>1$, the Brownian motion is transient, i.e., $\|W_t\|_2$ goes to infinity. How can I deal with this situation?
A sketch of a proof, based on the comment of Christian Remling at a now deleted MathOverflow question.
Let $x\in\mathbb{R}^d$ be any starting point, and define $S_1=\{y\in\mathbb{R}^d: \|y\|=2\|x\|\}$. Let $\tau_1$ be the hitting time of $S_1$ and $C$ be your cone. The function $z\mapsto\mathbb{P}_z(B_{\tau_1}\in C)$ is continuous and strictly positive for $z$ with $\|z\|=\|x\|$ and so, for such $z$, $$ \mathbb{P}_z(B_{\tau_1}\in C)\geq p>0.$$
Now define $S_n=\{y\in\mathbb{R}^d: \|y\|=2^n\|x\|\}$ so that, by scaling, $\mathbb{P}_z(B_{\tau_n}\in C)\geq p>0$ for $z$ with $\|z\|=2^{n-1}\|x\|$.
Using the strong Markov property gives $$\mathbb{P}_x(B_{\tau_n}\notin C\mbox{ for }n=1,\dots, N)\leq (1-p)^N,$$ and letting $N\to\infty$ we get $\mathbb{P}_x(\tau_C<\infty)=1.$