[Help-bash] Bash read specific element from csv

[Phil]

Hello,

I am struggling to find the bash code to read a specific element from a 
CSV.   The csv is constructed as follows:
<--------- SNAP --------->
"Naam";"ISIN";"Symbol";"Market";"Trading 
Currency";"Open";"Hoog";"Laag";"Last";"Last Date/Time";"Time 
Zone";"Volume";"Turnover"
"European Equities";"";;;;;;;;;;;
"28 Sep 2014";;;;;;;;;;;;
"All datapoints provided as of end of last active trading day.";;;;;;;;;;;;
"IDI";"FR0000051393";"IDIP";"Euronext 
Paris";"EUR";"23.42";"23.50";"23.42";"23.42";"26/09/2014 
12:00";"CET";"21";"492.98"
"BETER BED";"NL0000339703";"BBED";"Euronext 
Amsterdam";"EUR";"16.80";"16.80";"16.48";"16.645";"26/09/2014 
17:35";"CET";"1206";"20106.64"
"ALCATEL-LUCENT";"FR0000130007";"ALU";"Euronext 
Paris";"EUR";"2.494";"2.504";"2.446";"2.486";"26/09/2014 
17:39";"CET";"15581702";"38683091.21"
"FIPP";"FR0000038184";"FIPP";"Euronext 
Paris";"EUR";"0.08";"0.08";"0.08";"0.08";"26/09/2014 
10:35";"CET";"20001";"1600.08"
"ROBERTET CI";"FR0000045601";"CBE";"Euronext 
Paris";"EUR";"101.70";"101.70";"101.70";"101.70";"23/09/2014 
12:00";"CET";"40";"4068.00"
...
<--------- SNAP --------->

I want bash to read the first element on the third line, containing the 
date (ie 28 Sep 2014), and return into a string.

The closest I get is
> sed -n "3 p" myFile.csv
"28 Sep 2014";;;;;;;;;;;;
>

I thought that awk would help me out, but the following command give me only:
> sed -n "3 p" EU.csv | awk '{print $1}'
"28
>

Thanks in advance.
Best regards,

Phil