I don't understand this one small point about Brownian motion...

Here's the question, my answer and the part I always miss out...

Let $\displaystyle \{W_t\}_{t \geq 0}$ be a standard Brownian motion. Show that.

(a) $\displaystyle \{X_t\}_{t \geq0}$ is a standard Brownian motion, where $\displaystyle X_t = -W_t$

Question is easy enough. Can show that $\displaystyle X_0 = 0$ almost surely and get that for $\displaystyle 0 \leq s \leq t$, $\displaystyle X_t - X_s = -(W_t - W_s)$ is distributed with mean $\displaystyle 0$ and variance $\displaystyle t-s$ and is independent of $\displaystyle F_s = \sigma (W_u: 0 \leq u \leq s)$. (Not ENTIRELY sure why this is though)

Then comes this small statement that I don't know how to show (and hence don't know if it's true).

$\displaystyle X_t$ has continuous paths.

How do I show that? It's only ever stated in answers so I don't need some in depth proof, just a simple, it has continuous paths because...