The Wikipedia article on the Sierpinski carpet fractal says that it is compact, connected and locally connected.
It is clear from the construction that the Sierpinski carpet is closed and bounded in $\mathbb{R}^2$, hence it is compact. However, I don't see any reason for it to be connected and locally connected. I'm not necessarily asking for a rigorous proof; I'd be happy to see at least some kind of idea which would make me believe it is true.
If you look closely at the construction of Sierpiński carpet, you will notice that given a rational $q=\frac{m}{3^n}$ we have that both $(q,r)$ and $(s,q)$ belong to the Sierpiński carpet, for any $r,s\in [0,1]$. Meaning both $X$ and $Y$ cross sections at rationals with $3^n$ in denominator are full intervals.
And this means that any two such points can be connected by a path: (at most) two straight lines. Furthermore those points are dense in the carpet. So given a sequence of those points $v_n$ convergent to some point $v$ we can glue paths between $v_{n-1}$ and $v_n$ to obtain a path from $v_0$ to $v$. In general such construction does not have to yield a continuous path. However in this case it works, because we glue concrete paths: those two straight lines. Such path has the important property that the closer points are to each other the smaller the path is (meaning contained in a smaller ball). And this is enough to conclude that such construction gives a continuous path.
And so the carpet is even path connected. Similar argument works for local (path) connectedness.