Does the uniform convergence imply continuity? In a proof I am reading, the author claims that 
for $f(t)=\begin{cases}1\text{ if }t\in[\frac{2}{3},1]\\0\text{ if } t\in [0,\frac{1}{3}]\\3t-1\text{ if } (\frac{1}{3},\frac{2}{3})\end{cases}$
$x(t)=\sum_{k=0}^{\infty}2^{-k-1}f(3^{2k}t)$ and $y(t)=\sum_{k=0}^{\infty}2^{-k-1}f(3^{2k+1}t)$ is continuous by the Weierstrass M-Test, but I don't see why. Can anyone elaborate?

Yes, if $f_n\to f$ uniformly , $f_n$ is continuous $\forall n$, then $f$ is continuous. Here is the idea of proof.
In your example, $f$ is continuous and bounded, $x_n(t)=\sum_{k=0}^{n}2^{-k-1}f(3^{2k}t)$ is also continuous, and $x_n\to x$ uniformly by M-test, so $x$ is continuous.
