Why is this true: “There are countably many Turing Machines”
A quick informal answer:


a Turing Machine (states, transitions, ecc.) can be encoded using a string of $0$'s and $1$'s;
so you can pick all the binary strings in lexicographic order (0,1,00,01,10,11,000,001,...) and enumerate the Turing machines (i.e. build a one to one corrispondence between natural numbers and Turing machines) repeating the following steps:

1) start with $i=1$, $m=1$
2) generate the next binary string $s_i$ in lexicographic order
3.1) if $s_i$ is a valid encoding of a Turing machine then output $s_i$ as the $m$-th Turing machine and set $m = m+1$;
3.2) if $s_i$ is not a valid encoding of a Turing machine then ignore it
4) set $i = i+1$ and goto step 2


In this way each natural number ($m= 1,2,3,...$) corresponds to a Turing machine, and each Turing machine has a corresponding $m$, because you scan all possible binary strings. So the set of Turing machines is countable.

On the other side consider all possible strings $S$ over alphabet $\{0,1\}$:
$S =\{0,1\}^* = \{\epsilon,0,1,00,01,10,11,000,001,...\}$

A language $L$ is a subset (possible infinite) of $S$: $L \subseteq S$.

So the set of all languages is exactly the power set of $S$:
$2^S = \{ \{\epsilon\}, \{0\}, \{1\}, \{0,1\}, ... \}$ 

But the power set of a countably infinite set is uncountable (it can be easily proved using the diagonalization method).

You can also apply the diagonalization method directly to the set of languages: suppose that the languages are countable, then we can arrange all of them in a table in which every (infinte) row $i$ represents the elements of the language $L_i$ and the columns represent the string over the alphabet $\{0,1\}$ ($(i,j)=1$ if and only if string $j$ is in $L_i$):

    0 1 00 01 10 11 ... 
L1  0 0  1  0  1  0
L2  1 1  0  1  0  1
L3  0 1  0  0  1  0
...


Then define a new language changing the "membership flag" of the elements $(i,i)$:

    0  1 00 01 10 11 ... 
L1 [1] 0  1  0  1  0
L2  1 [0] 0  1  0  1
L3  0  1 [1] 0  1  0
...


The new language $L_{new} = \{0,00,...\}$ built from the modified diagonal is different from every language $L_i$: $0 \notin L_1, 0 \in L_{new}$, $1 \in L_2, 1 \notin L_{new}$, $00 \notin L_3, 00 \in L_{new}$, .... But this is a contraddiction because by hypothesis the above table should be an enumeration of all languages.