The ideal Monte Carlo algorithm uses independent successive random values. In MCMC, successive values are not independant, which makes the method converge slower than ideal Monte Carlo; however, the faster it mixes, the faster the dependence decays in successive iterations¹, and the faster it converges.
¹ I mean here that the successive values are quickly "almost independent" of the initial state, or rather that given the value $X_n$ at one point, the values $X_{ń+k}$ become quickly "almost independent" of $X_n$ as $k$ grows; so, as qkhhly says in the comments, "the chain don’t keep stuck in a certain region of the state space".
Edit: I think the following example can help
Imagine you want to estimate the mean of the uniform distribution on $\{1, \dots, n\}$ by MCMC. You start with the ordered sequence $(1, \dots, n)$; at each step, you chose $k>2$ elements in the sequence and randomly shuffle them. At each step, the element at position 1 is recorded; this converges to the uniform distribution. The value of $k$ controls the mixing rapidity: when $k=2$, it is slow; when $k=n$, the successive elements are independent and the mixing is fast.
Here is a R function for this MCMC algorithm :
mcmc <- function(n, k = 2, N = 5000)
{
x <- 1:n;
res <- numeric(N)
for(i in 1:N)
{
swap <- sample(1:n, k)
x[swap] <- sample(x[swap],k);
res[i] <- x[1];
}
return(res);
}
Let’s apply it for $n = 99$, and plot the successive estimation of the mean $\mu = 50$ along the MCMC iterations:
n <- 99; mu <- sum(1:n)/n;
mcmc(n) -> r1
plot(cumsum(r1)/1:length(r1), type="l", ylim=c(0,n), ylab="mean")
abline(mu,0,lty=2)
mcmc(n,round(n/2)) -> r2
lines(1:length(r2), cumsum(r2)/1:length(r2), col="blue")
mcmc(n,n) -> r3
lines(1:length(r3), cumsum(r3)/1:length(r3), col="red")
legend("topleft", c("k = 2", paste("k =",round(n/2)), paste("k =",n)), col=c("black","blue","red"), lwd=1)

You can see here that for $k=2$ (in black), the convergence is slow; for $k=50$ (in blue), it is faster, but still slower than with $k=99$ (in red).
You can also plot an histogram for the distribution of the estimated mean after a fixed number of iterations, eg 100 iterations:
K <- 5000;
M1 <- numeric(K)
M2 <- numeric(K)
M3 <- numeric(K)
for(i in 1:K)
{
M1[i] <- mean(mcmc(n,2,100));
M2[i] <- mean(mcmc(n,round(n/2),100));
M3[i] <- mean(mcmc(n,n,100));
}
dev.new()
par(mfrow=c(3,1))
hist(M1, xlim=c(0,n), freq=FALSE)
hist(M2, xlim=c(0,n), freq=FALSE)
hist(M3, xlim=c(0,n), freq=FALSE)

You can see that with $k=2$ (M1), the influence of the initial value after 100 iterations only gives you a terrible result. With $k=50$ it seems ok, with still greater standard deviation than with $k=99$. Here are the means and sd:
> mean(M1)
[1] 19.046
> mean(M2)
[1] 49.51611
> mean(M3)
[1] 50.09301
> sd(M2)
[1] 5.013053
> sd(M3)
[1] 2.829185