To reproduce in Python 3.4.2:

def f():
    x = None
    def g():
        nonlocal x
        try:
            raise Exception()
        except Exception as x:
            pass
    g()
    # ↓ UnboundLocalError: local variable 'x' referenced before assignment
    print("x", x)
f()

Compare this to:

def f():
    x = None
    def g():
        nonlocal x
        with open("/dev/null") as x:
            pass
    g()
    print("x", x)
f()

(which prints: x <_io.TextIOWrapper name='/dev/null' mode='r' encoding='UTF-8'>)

The first example seems to behave as I would expect. The UnboundLocalError is raised by the print() call, because the “x” variable has been deleted by the exception handler. Equivalent code without using “nonlocal”:

>>> def f():
...     x = None
...     try:
...         raise Exception()
...     except Exception as x:
...         pass
...     print("x", x)  # UnboundLocal due to exception handler
... 
>>> f()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 7, in f
UnboundLocalError: local variable 'x' referenced before assignment

In both cases, I think this is correct behaviour. See <https://docs.python.org/3.4/reference/compound_stmts.html#except>, which says “When an exception has been assigned using ‘as target’, it is cleared at the end of the except clause.”

Indeed, if you replace the except clause with a 'del x', you get the same UnboundLocalError.  This is working as designed.