Comments and demonstration:  Of the several pages linked in previous comments, perhaps [this one](https://stats.stackexchange.com/questions/193048/how-to-get-the-maximum-likelihood-estimator-of-u-theta-theta-1)
from @stubbornAtom and 
[this one](https://stats.stackexchange.com/questions/360725/finding-maximum-likelihood-estimator-symmetric-uniform-distribution) from @whuber are most directly relevant.

While there is no unique MLE, it makes sense to use sufficient statistics to find an unbiased estimator with small root mean
square error (RMSE). One good candidate is the unbiased midrange, as illustrated in the R simulation below for the case in which
$\theta = 5.$

    set.seed(311)
    m = 10^6; mx = mn = mr = numeric(m)
    th = 5; n = 10
    for(i in 1:m) {
      x = runif(n, th, th+1)
      mx[i] = max(x)
      mn[i] = min(x)
      mr[i] = mean(range(x)) }

.

    # Unbiased estimators
    t1 = mx - .9    # adj max
    t2 = mr - .5    # adj midrange
    t3 = mn - .1    # adj min

    mean(t1); mean(t2);  mean(t3)
    [1] 5.009175
    [1] 5.00002
    [1] 4.990866
    sd(t1); sd(t2); sd(t3)
    [1] 0.08292119
    [1] 0.06145805            # smallest of 3
    [1] 0.08282185

    # RMSE's
    sqrt(mean((t1 -th)^2))
    [1] 0.08342722
    sqrt(mean((t2 -th)^2))
    [1] 0.06145803            # smallest of 3
    sqrt(mean((t3 -th)^2))
    [1] 0.08332399

    par(mfrow=c(1,3))
     hist(t1, prob=T, col="skyblue2", main="Unbiased Max")
     hist(t2, prob=T, col="skyblue2", main="Unbiased Midrange")
     hist(t3, prob=T, col="skyblue2", main="Unbiased Min")
    par(mfrow=c(1,3))

[![enter image description here][2]][2]


  [1]: https://i.sstatic.net/kbybv.png
  [2]: https://i.sstatic.net/NNCzz.png